Problema que le pongo a Qwen y su solución

 Ha pensado mucho y pongo sus pensamientos, traducidos al español por deepseek

ENUNCIADO.

Considera la función compleja de variable compleja que a cada número complejo le hace corresponder la raiz cuadrada de z (rama principal) si la parte imaginaria de z es mayor o igual que cero y el opuesto a la raíz cuadrada de z (rama principal) si la parte imaginaria de z es menor que cero.

a) Escribe la expresión de la función en lenguaje matemático.

b) Halla el dominio de la función

c) Halla el codominio o conjunto imagen de la función.

d) ¿Para qué valores de z es continua?

Resuelve paso a paso y da explicaciones detalladas de cada paso

PENSAMIENTO TRADUCIDO AL ESPAÑOL

Bien, abordemos este problema complejo sobre funciones. La pregunta trata sobre definir una función que mapee cada número complejo $z$ a su raíz cuadrada principal o a su negativo, dependiendo de la parte imaginaria de $z$. Analicemos esto paso a paso.

Parte (a): Necesitamos escribir la expresión matemática para esta función. La raíz cuadrada principal de un número complejo $z$ se define usualmente con un corte de rama a lo largo del eje real negativo, ¿correcto? Así, el valor principal tiene argumentos entre $-\pi$ y $\pi$. Pero aquí especifican que si la parte imaginaria de $z$ es no negativa, tomamos la raíz cuadrada principal, y si es negativa, tomamos el negativo de la raíz cuadrada principal. Hmm, parece que estamos dividiendo el plano complejo a lo largo del eje real en lugar del corte de rama usual. Espera, pero la rama principal normalmente se define con un corte de rama en el eje real negativo. ¿Quizás debamos ajustar cómo definimos las raíces aquí?

Recuerdo que para la raíz cuadrada principal, dado $z=r{e}^{i\theta }$ con $-\pi <\theta \le \pi$, entonces $\sqrt{z}=\sqrt{r}{e}^{i\theta \mathrm{/}2}$. Si $\mathrm{Im}(z)\ge 0$, eso implica $\theta \in [0,\pi ]$, y la raíz cuadrada tendría $\theta \mathrm{/}2\in [0,\pi \mathrm{/}2]$. Si $\mathrm{Im}(z)<0$, entonces $\theta \in (-\pi ,0)$, y la raíz cuadrada tendría $\theta \mathrm{/}2\in (-\pi \mathrm{/}2,0)$. Pero en este problema, cuando $\mathrm{Im}(z)<0$, tomamos el negativo de la raíz cuadrada principal. Calculemos eso: el negativo de $\sqrt{z}$ sería $-\sqrt{r}{e}^{i\theta \mathrm{/}2}$. Si $\theta$ es negativo, digamos $\theta =-\varphi$ con $\varphi >0$, entonces $\theta \mathrm{/}2=-\varphi \mathrm{/}2$, así $e^{i\theta \mathrm{/}2}=e^{-i\varphi \mathrm{/}2}$, y multiplicar por $-1$ da $-e^{-i\varphi \mathrm{/}2}=e^{i(\pi -\varphi \mathrm{/}2)}$. ¿Acaso esta raíz negativa podría reinterpretarse como otra rama?

Alternativamente, quizás esta función busca que las raíces sean continuas a través del eje real, excepto quizás en el origen. Normalmente, la rama principal tiene una discontinuidad al cruzar el eje real negativo. Pero aquí, al dividir la definición según el signo de la parte imaginaria, ¿intentan lograr continuidad en el eje real?

Para la parte (a), solo debo escribir la expresión. Denotemos la función como $f(z)$. Entonces:

$$f(z)=\{\begin{array}{ll}\sqrt{z}(\text{principal})& \text{si }\mathrm{Im}(z)\ge 0,\\ -\sqrt{z}(\text{principal})& \text{si }\mathrm{Im}(z)<0.\end{array}$$

¿Cómo expresar $\sqrt{z}$ en términos de $z$? En forma polar: si $z\mathrm{\ne }0$, escribimos $z=r{e}^{i\theta }$, con $r\ge 0$ y $\theta \in (-\pi ,\pi ]$. Entonces $\sqrt{z}=\sqrt{r}{e}^{i\theta \mathrm{/}2}$. Así, $f(z)$ sería $\sqrt{r}{e}^{i\theta \mathrm{/}2}$ cuando $\mathrm{Im}(z)\ge 0$ (correspondiente a $\theta \in [0,\pi ]$), y $-\sqrt{r}{e}^{i\theta \mathrm{/}2}$ cuando $\theta \in (-\pi ,0)$ (i.e., $\mathrm{Im}(z)<0$). Nota: al multiplicar por $-1$, reflejamos el resultado al semiplano superior.

Parte (b): El dominio de la función. La raíz cuadrada principal se define en todos los complejos excepto en el corte de rama. Pero como $f(z)$ combina $\sqrt{z}$ y $-\sqrt{z}$, su dominio es $\mathbb{C}$, ya que incluso en el eje real negativo ($\mathrm{Im}(z)=0$), $\sqrt{z}$ está definida (p.ej., $\sqrt{-1}=i$).

Parte (c): Imagen o codominio. Al analizar casos, $f(z)$ mapea todo $z$ al semiplano superior cerrado $\{w\in \mathbb{C}\mid \mathrm{Im}(w)\ge 0\}$. Esto incluye:

• Raíces en el primer cuadrante para $\mathrm{Im}(z)\ge 0$.

• Negativos de raíces en el cuarto cuadrante (segundo cuadrante tras reflejar) para $\mathrm{Im}(z)<0$.

Parte (d): Continuidad. La función es continua en todo $\mathbb{C}$ excepto en el eje real positivo, donde los límites por arriba y abajo difieren (p.ej., $\sqrt{z}$ vs. $-\sqrt{z}$). En el eje real negativo, los límites coinciden (ambos tienden a $i\sqrt{\mathrm{\mid }z\mathrm{\mid }}$), garantizando continuidad allí.

PENSAMIENTO DE QWEN ORIGINAL EN INGLÉS, PORQUE CREO QUE DEEPSEEK HA HECHO UNA TRADUCCIÓN MU RESUMIDA A PESAR DE QUE LE HE PEDIDO QUE SEA FIEL AL TEXTO

Okay, let's tackle this complex function problem. So, the question is about defining a function that maps each complex number z to either the principal square root of z or its negative, depending on the imaginary part of z. Let me break this down step by step. Starting with part (a): We need to write the mathematical expression for this function. The principal square root of a complex number z is usually defined with a branch cut along the negative real axis, right? So the principal value has arguments between -π and π. But here, they specify that if the imaginary part of z is non-negative, we take the principal square root, and if it's negative, we take the negative of the principal square root. Hmm, that seems like we're splitting the complex plane along the real axis instead of the usual branch cut. Wait, but the principal branch is usually defined with a branch cut on the negative real axis. So maybe we need to adjust how we define the square roots here? Let me recall that for the principal square root, given z = r e^{iθ} where -π < θ ≤ π, then sqrt(z) = sqrt(r) e^{iθ/2}. So if Im(z) ≥ 0, that means θ is between 0 and π, and the square root would have θ/2 between 0 and π/2. If Im(z) < 0, then θ is between -π and 0, and the square root would have θ/2 between -π/2 and 0. But in this problem, when Im(z) < 0, we take the negative of the principal square root. Let me compute that: negative of sqrt(z) would be -sqrt(r) e^{iθ/2}. But if θ is negative, say θ = -φ where φ > 0, then θ/2 = -φ/2, so e^{iθ/2} = e^{-iφ/2}, and multiplying by -1 gives -e^{-iφ/2} = e^{i(π - φ/2)}. Wait, so maybe this negative of the principal square root can be reinterpreted as another branch? Alternatively, perhaps this function is designed to make the square roots continuous across the real axis except maybe at the origin? Let me think. Normally, the principal branch has a discontinuity when crossing the negative real axis. But here, by splitting the definition based on the sign of the imaginary part, maybe they are trying to have continuity across the real axis? But for part (a), I just need to write the expression. Let me denote the function as f(z). Then: f(z) = { sqrt(z) (principal) if Im(z) ≥ 0, -sqrt(z) (principal) if Im(z) < 0 } But how to write sqrt(z) in terms of z? Well, using polar form. If z ≠ 0, we can write z = r e^{iθ}, where r ≥ 0 and θ ∈ (-π, π]. Then sqrt(z) is sqrt(r) e^{iθ/2}. So the function f(z) would be sqrt(r) e^{iθ/2} when Im(z) ≥ 0, which corresponds to θ ∈ [0, π], and -sqrt(r) e^{iθ/2} when θ ∈ (-π, 0), which is Im(z) < 0. Wait, but if θ is negative, then θ/2 is negative, so e^{iθ/2} is in the lower half-plane. Multiplying by -1 would reflect it to the upper half-plane? Let me check: Suppose θ = -φ, where 0 < φ < π. Then sqrt(z) = sqrt(r) e^{-iφ/2}, which is in the lower half-plane. Then -sqrt(z) = -sqrt(r) e^{-iφ/2} = sqrt(r) e^{i(π - φ/2)}, which is in the upper half-plane. Interesting. So effectively, this function f(z) maps both the upper and lower half-planes to the upper half-plane? Or does it? Wait, let's take an example. Let z = -1, which has Im(z) = 0. Wait, the problem says Im(z) ≥ 0, so Im(z)=0 is included in the first case. For z = -1, θ = π, so sqrt(z) = i. But Im(z)=0 here? Wait, no: z = -1 is on the real axis, so Im(z)=0. Therefore, according to the definition, we take the principal square root. Which is i. Now take z = 1, which is also on the real axis with Im(z)=0, so sqrt(1) = 1. Now take z = i (Im(z)=1 ≥0), so f(z) = sqrt(i) = e^{iπ/4} = (sqrt(2)/2 + i sqrt(2)/2). If z = -i (Im(z) = -1 < 0), then f(z) = -sqrt(-i). Let's compute sqrt(-i). The principal sqrt of -i: since -i has θ = -π/2, so sqrt(-i) = e^{-iπ/4} = cos(-π/4) + i sin(-π/4) = sqrt(2)/2 - i sqrt(2)/2. Then -sqrt(-i) = -sqrt(2)/2 + i sqrt(2)/2, which is e^{i3π/4}, which is in the upper half-plane. So regardless of whether z is in the upper or lower half-plane, f(z) ends up in the upper half-plane? Except maybe on the real axis. Wait, so codomain might be the upper half-plane union the positive real axis? Let me check. Wait, for z real positive (Im(z)=0), f(z) = sqrt(z) which is positive real. For z real negative (Im(z)=0), sqrt(z) is i*sqrt(|z|), which is purely imaginary positive. Wait, wait z real negative: z = -r, r>0. Then sqrt(z) is sqrt(r) e^{iπ/2} = i sqrt(r). So yes, the principal square root of a negative real number is purely imaginary with positive imaginary part. Therefore, for z real negative, f(z) = i sqrt(|z|), which is in the upper half-plane. Wait, so for all z ≠ 0, f(z) is in the upper half-plane? Because when Im(z) ≥0, sqrt(z) principal has argument θ/2 with θ ∈ [0, π], so θ/2 ∈ [0, π/2], which is the first quadrant. Wait, no: θ ∈ (-π, π]. If z is in the lower half-plane (Im(z) <0), θ ∈ (-π, 0), then sqrt(z) principal is e^{iθ/2} with θ/2 ∈ (-π/2, 0), which is in the fourth quadrant. Then taking negative of that sqrt(z) would give -sqrt(r) e^{iθ/2} = sqrt(r) e^{i(θ/2 + π)}, which would be θ/2 + π. Since θ/2 was in (-π/2, 0), adding π gives (π/2, π). So arguments between π/2 and π, which is the second quadrant. Therefore, combining both cases, f(z) has arguments in [0, π/2] ∪ [π/2, π] = [0, π], so the entire upper half-plane. However, note that when θ approaches π from below (upper half-plane), then θ/2 approaches π/2, and when θ approaches -π from above (lower half-plane), then θ/2 approaches -π/2, but taking -sqrt(z) gives an argument of -π/2 + π = π/2. Wait, so approaching the negative real axis from above or below would both give f(z) approaching the positive imaginary axis? Wait, let's take z approaching -1 from above (slightly positive imaginary part) and from below (slightly negative imaginary part). If z approaches -1 from above (Im(z) → 0+), then f(z) = sqrt(z) approaches i. If z approaches -1 from below (Im(z) → 0-), then f(z) = -sqrt(z). Let's compute sqrt(z) near -1 from below: z = -1 - εi, where ε → 0+. The principal sqrt would have θ approaching π from below, so θ ≈ π - δ, so θ/2 ≈ π/2 - δ/2. Wait, but if z is approaching -1 from below, maybe θ approaches -π? Wait, no. The principal value of θ for z approaching -1 from above is θ approaching π, and from below, θ approaching -π. So sqrt(z) when approaching from above would be sqrt(r) e^{iπ/2} = i, and from below, sqrt(z) would be sqrt(r) e^{-iπ/2} = -i. Then -sqrt(z) would be i. Therefore, both sides approach i. Therefore, the function f(z) is continuous across the negative real axis? Interesting. Whereas the principal square root has a branch cut on the negative real axis because approaching from above and below gives different limits. But here, because on the lower half-plane we take the negative, the limits agree on the negative real axis. However, on the positive real axis, what happens? If z approaches a positive real number from above (Im(z) → 0+) and from below (Im(z) → 0-). For z approaching from above, f(z) = sqrt(z) which approaches the positive real number. For z approaching from below, f(z) = -sqrt(z). But sqrt(z) here is the principal sqrt, which for positive real numbers is just the positive real root. Therefore, approaching from below, f(z) approaches -sqrt(z), which is the negative real number. Therefore, there's a discontinuity on the positive real axis? Because approaching from above gives positive sqrt, from below gives negative sqrt. So this function f(z) is continuous everywhere except on the positive real axis? Wait, but the problem asks in part (d) where it's continuous. But back to part (a). So writing the function in mathematical terms. So f(z) is defined as follows: For z ∈ ℂ \ {0}, write z = r e^{iθ} with r > 0 and θ ∈ (-π, π]. Then: f(z) = sqrt(r) e^{iθ/2} if Im(z) ≥ 0 (which corresponds to θ ∈ [0, π]), and f(z) = -sqrt(r) e^{iθ/2} if Im(z) < 0 (which corresponds to θ ∈ (-π, 0)). But since θ ∈ (-π, 0) for Im(z) < 0, then θ/2 ∈ (-π/2, 0), so sqrt(r) e^{iθ/2} is in the fourth quadrant, and multiplying by -1 reflects it to the second quadrant (arguments between π/2 and π). Alternatively, we can express the function as: f(z) = sqrt(z) if Im(z) ≥ 0, f(z) = -sqrt(z) if Im(z) < 0. But we have to be careful with the definition of sqrt(z) here. Since sqrt(z) is the principal branch, which has the branch cut on the negative real axis. However, in our case, the function f(z) combines sqrt(z) and -sqrt(z) depending on the sign of Im(z). Therefore, the mathematical expression is simply: f(z) = { sqrt(z) if Im(z) ≥ 0, -sqrt(z) if Im(z) < 0 } Where sqrt(z) denotes the principal square root. Alternatively, using the polar form: f(z) = sqrt(|z|) e^{i arg(z)/2} if Im(z) ≥ 0, f(z) = -sqrt(|z|) e^{i arg(z)/2} if Im(z) < 0. But since arg(z) for Im(z) < 0 is in (-π, 0), so maybe we can write it as: f(z) = sqrt(|z|) e^{i arg(z)/2} for Im(z) ≥ 0, f(z) = -sqrt(|z|) e^{i arg(z)/2} for Im(z) < 0. Alternatively, since for Im(z) < 0, we can write arg(z) = -arg(\overline{z}), but not sure if that helps. Alternatively, notice that when Im(z) < 0, then f(z) = -sqrt(z) where sqrt(z) is principal. But sqrt(\overline{z}) = \overline{sqrt(z)}, so maybe f(z) relates to conjugation? Not sure. But for part (a), probably the simplest way is to write: f(z) = \begin{cases} \sqrt{z} & \text{if } \Im(z) \geq 0, \\ -\sqrt{z} & \text{if } \Im(z) < 0. \end{cases} With sqrt(z) being the principal branch. However, sometimes the principal branch is defined with a branch cut on the negative real axis, so we might need to specify that sqrt(z) is the principal branch, i.e., with argument in (-π, π]. But maybe that's understood. Moving to part (b): the domain of the function. Well, the square root function is defined for all complex numbers except the branch cut. But since we're using the principal branch, which excludes z ≤ 0 on the real axis? Wait, no: the principal square root is defined on all complex numbers except the negative real axis (including zero?), but actually sqrt(0) = 0. So domain is ℂ \ (-∞, 0]? Wait, but in our function f(z), we have sqrt(z) for Im(z) ≥ 0 and -sqrt(z) for Im(z) < 0. However, points on the real axis with Im(z)=0: for z > 0, Im(z)=0, so included in the first case; for z ≤ 0, Im(z)=0, still included in the first case. Wait, the definition says Im(z) ≥ 0, so Re(z) can be anything as long as Im(z) is non-negative. Therefore, the domain is all complex numbers except where? Well, sqrt(z) is defined for all z ≠ 0? Wait, no: sqrt(z) is defined for all z ≠ 0, but with a branch cut. Wait, actually, the principal square root is defined on all complex numbers except the branch cut, which is typically the negative real axis. Therefore, the domain is ℂ \ (-∞, 0], but in our function, we are defining f(z) for all z, because even if z is on the negative real axis, since Im(z)=0, we take the principal sqrt(z), which is defined there. Wait, but sqrt(z) for z negative real is defined as i*sqrt(|z|), right? So technically, the principal square root is defined for all z ∈ ℂ, with a branch cut on (-∞, 0], but the function is defined on all z, but discontinuous across the negative real axis. Wait, no: actually, the principal value is defined on ℂ \ (-∞, 0], because at the branch cut, it's discontinuous but still defined on one side. Wait, no: the principal square root is defined for all z ≠ 0 by taking θ in (-π, π], but at z negative real, approaching from above and below gives different limits. However, as a function, it's defined on all z ≠ 0, with a branch cut on the negative real axis where it's discontinuous. So technically, the domain of sqrt(z) is ℂ \ {0}, but it's discontinuous on (-∞, 0). Therefore, our function f(z) uses sqrt(z) for all z, so its domain is ℂ \ {0}? Wait, but z=0: sqrt(0) = 0, so f(0) would be sqrt(0) since Im(0)=0 ≥0, so f(0)=0. Therefore, the domain is actually all of ℂ, because even at z=0 it's defined, and for other z, sqrt(z) is defined (even though it has a branch cut). Wait, but the principal square root function is usually considered to have domain ℂ \ (-∞, 0], because at z=0 it's defined. Wait, no: sqrt(z) at z=0 is 0, so it's fine. So the domain is all complex numbers. Because even though there's a branch cut on (-∞, 0], the function is still defined there, just discontinuous. So in complex analysis, the domain of the principal square root is ℂ, because it's defined everywhere, but it's not analytic on (-∞, 0]. Therefore, our function f(z) is defined for all z ∈ ℂ. Because for any complex number, we can compute sqrt(z) (principal branch) and then apply the sign based on Im(z). Therefore, the domain is ℂ. But let me confirm. For z=0, f(z) = sqrt(0) = 0, which is fine. For any other z, even on the negative real axis, sqrt(z) is defined. Therefore, domain is all complex numbers. So answer to (b) is ℂ. Part (c): Codomain or image set. Earlier, I thought that f(z) maps everything to the upper half-plane. Let's verify. Take any z ≠ 0. If Im(z) ≥ 0, then f(z) = sqrt(z), which in polar coordinates has argument θ/2, θ ∈ [0, π], so θ/2 ∈ [0, π/2]. So these are complex numbers in the first quadrant (including the positive real and positive imaginary axes). If Im(z) < 0, then f(z) = -sqrt(z). sqrt(z) here has θ ∈ (-π, 0), so θ/2 ∈ (-π/2, 0), which is the fourth quadrant. Multiplying by -1 reflects it to the second quadrant (arguments between π/2 and π). Therefore, combining both cases, f(z) maps to the upper half-plane (Im(w) ≥ 0) union... Wait, the first quadrant has Im(w) ≥ 0 and Re(w) ≥ 0, while the second quadrant has Im(w) ≥ 0 and Re(w) ≤ 0. So together, all complex numbers with non-negative imaginary part, except... Wait, does every complex number with Im(w) > 0 get covered? Yes. Because given any w with Im(w) > 0, we can write w^2 = z. If w is in the first quadrant, then z has Im(z) ≥ 0; if w is in the second quadrant, then z has Im(z) < 0. Wait, let's see: If w is in the first quadrant, w = a + ib with a > 0, b > 0, then z = w² = (a² - b²) + i2ab. The imaginary part is positive, so z is in the upper half-plane. Then f(z) = sqrt(z) = w. Wait, but sqrt(z) could be ±w, but since we take the principal branch, which is in the first quadrant. Similarly, if w is in the second quadrant, w = -a + ib with a > 0, b > 0, then z = w² = (a² - b² - 2iab)? Wait, no: Wait, let's compute w = c e^{iφ}, φ ∈ (0, π). Then w² = c² e^{i2φ}. If φ ∈ (0, π/2), then 2φ ∈ (0, π), so Im(z) > 0. If φ ∈ (π/2, π), then 2φ ∈ (π, 2π), so Im(z) < 0 when 2φ ∈ (π, 2π), i.e., φ ∈ (π/2, π). Therefore, w in second quadrant (φ ∈ (π/2, π)) maps to z with Im(z) < 0. Therefore, for such z, f(z) = -sqrt(z). But sqrt(z) would be in the fourth quadrant (since z has argument in (π, 2π), principal sqrt would have argument in (π/2, π), wait: Wait z has argument θ = 2φ ∈ (π, 2π) since φ ∈ (π/2, π). Then sqrt(z) is sqrt(|z|) e^{iθ/2} = sqrt(|z|) e^{iφ}, which is w. Wait, but θ is in (π, 2π), so θ/2 is in (π/2, π), so sqrt(z) is in the second quadrant. Therefore, f(z) = -sqrt(z) would be in the fourth quadrant? Wait, no: Wait, if sqrt(z) = w' = -a + ib (second quadrant), then -sqrt(z) = a - ib (fourth quadrant). But according to our earlier analysis, f(z) when Im(z) <0 should be in the second quadrant. There's a contradiction here. Let me re-examine. Wait, take w in the second quadrant: w = -a + ib, a,b >0. Let's compute z = w². z = (-a + ib)^2 = a² - 2iab - b² = (a² - b²) - i2ab. So Re(z) = a² - b², Im(z) = -2ab < 0. So z is in the lower half-plane. Therefore, f(z) = -sqrt(z). What's sqrt(z)? Given z = (a² - b²) - i2ab. Let's compute its modulus and argument. |z| = sqrt( (a² - b²)^2 + (2ab)^2 ) = sqrt(a^4 - 2a²b² + b^4 + 4a²b²) = sqrt(a^4 + 2a²b² + b^4) = (a² + b²). The argument θ of z is arctan( (-2ab)/(a² - b²) ). Depending on the values of a and b, this could be in different quadrants. But since Im(z) < 0, θ ∈ (-π/2, 0) if Re(z) >0, or θ ∈ (-π, -π/2) if Re(z) <0. Wait, Re(z) = a² - b². So if a > b, Re(z) >0; if a < b, Re(z) <0; if a = b, Re(z)=0. Case 1: a > b. Then Re(z) >0, so z is in the fourth quadrant. Its argument θ is in (-π/2, 0). Then sqrt(z) has argument θ/2 ∈ (-π/4, 0), which is in the fourth quadrant. Then -sqrt(z) is in the second quadrant. Case 2: a < b. Then Re(z) <0, so z is in the third quadrant (Im(z) <0, Re(z) <0). Its argument θ is in (-π, -π/2). Then sqrt(z) has argument θ/2 ∈ (-π/2, -π/4), which is in the fourth quadrant. Then -sqrt(z) is in the second quadrant. Case 3: a = b. Then Re(z)=0, so z = -i2a², which is on the negative imaginary axis. Its argument is -π/2. Then sqrt(z) has argument -π/4, so sqrt(z) = sqrt(2a²) e^{-iπ/4} = a√2 (cos(-π/4) + i sin(-π/4)) = a√2 (√2/2 - i√2/2) = a(1 - i). Then -sqrt(z) = -a + ai, which is in the second quadrant. Therefore, in all cases, f(z) = -sqrt(z) for z in lower half-plane maps to the second quadrant or the positive imaginary axis. But earlier, when we considered w in second quadrant, z = w² maps to lower half-plane, and f(z) = -sqrt(z). But sqrt(z) = w', some complex number in fourth quadrant, so -sqrt(z) is in second quadrant. However, z = w² for w in second quadrant, but how does f(z) relate to w? Wait, suppose we want to find for a given w in the upper half-plane (Im(w) >0), is there a z such that f(z) = w? If w is in the first quadrant: yes, take z = w², which has Im(z) ≥0, so f(z) = sqrt(z) = w. If w is in the second quadrant: take z = ( -w )² = w², since (-w)² = w². Wait, z = w² has Im(z) = Im( (a + ib)^2 ) = 2ab, but if w is in second quadrant, w = -c + id, c,d >0, then w² = c² - d² - i2cd. So Im(z) = -2cd <0. Therefore, z is in lower half-plane, so f(z) = -sqrt(z). Now sqrt(z) would be sqrt(w²) but in principal branch. Wait, z = w², but sqrt(z) could be w or -w depending on the branch. Wait, z = w², w in second quadrant. The principal sqrt(z) would have argument θ/2 where θ is the argument of z. Since z has argument in (-π, -π/2) if Re(z) <0 (as in case 2 above), or in (-π/2, 0) if Re(z) >0. Either way, sqrt(z) is in the fourth quadrant. But w is in second quadrant, so sqrt(z) ≠ w. Instead, sqrt(z) = conjugate(w)/something? Wait, this is getting confusing. Let's take a concrete example. Let w = i (which is on the positive imaginary axis). Then w² = -1. Then z = -1, which has Im(z)=0, so f(z) = sqrt(-1) = i. So f(-1) = i. Good. Another example: w = e^{iπ/4} (sqrt(i)), which is in first quadrant. Then z = w² = i, which has Im(z)=1 ≥0, so f(z)=sqrt(i)=w. Correct. Take w = e^{i3π/4} (in second quadrant). Then z = w² = e^{i3π/2} = -i. Then Im(z) = -1 <0, so f(z) = -sqrt(-i). Compute sqrt(-i): as before, sqrt(-i) = e^{-iπ/4}, so -sqrt(-i) = e^{i3π/4} = w. Perfect, so f(z) = w. Therefore, for any w in the upper half-plane (including the real axis?), wait Im(w) ≥0? Wait w = real positive: w = a >0. Then z = a², Im(z)=0, so f(z)=sqrt(a²)=a=w. w =0: z=0, f(z)=0. w real negative: w = -a <0. Can we find z such that f(z)=w? Suppose so. Then since f(z) is either sqrt(z) (upper half-plane or real positive) or -sqrt(z) (which for z in lower half-plane gives upper half-plane). Wait, according to previous examples, f(z) always lies in the closed upper half-plane (Im(w) ≥0). Because sqrt(z) when Im(z) ≥0 has Im(f(z)) ≥0, and -sqrt(z) when Im(z) <0 also results in Im(f(z)) ≥0 as shown earlier. Therefore, the image of f is the closed upper half-plane, including the real axis. But wait, when z is in the lower half-plane, f(z) = -sqrt(z). sqrt(z) is in the fourth quadrant, so -sqrt(z) is in the second quadrant (Im >0). When z is in upper half-plane, f(z)=sqrt(z) is in first quadrant (Im ≥0). On the real axis: z ≥0, f(z)=sqrt(z) real ≥0; z ≤0, f(z)=sqrt(z) purely imaginary positive. Therefore, the image is all complex numbers with Im(w) ≥0. So codomain is the closed upper half-plane, { w ∈ ℂ | Im(w) ≥0 }. But let's verify if every point in the closed upper half-plane is attained. Take any w with Im(w) ≥0. If w is in the first quadrant or on the positive real/imaginary axes, set z = w², which has Im(z) ≥0, so f(z)=sqrt(z)=w. If w is in the second quadrant, define z = w². Since w is in second quadrant (argument between π/2 and π), z = w² has argument between π and 2π, so Im(z) <0. Therefore, f(z) = -sqrt(z). But sqrt(z) should be w', where w'² = z and w' is in the principal branch, i.e., in the first quadrant? Wait no: Wait z = w², with w in second quadrant: z has argument 2φ where φ ∈ (π/2, π), so 2φ ∈ (π, 2π). So sqrt(z) is defined as having argument φ (half of 2φ), which is in (π/2, π), i.e., the second quadrant. Wait, but the principal branch sqrt(z) has arguments in (-π/2, π/2], right? No! Wait, confusion arises here. Hold on, the principal branch of sqrt(z) has arguments θ/2 where θ ∈ (-π, π]. Therefore, θ/2 ∈ (-π/2, π/2]. So sqrt(z) always lands in the closed right half-plane (Re ≥0). Wait, that contradicts previous statements. Wait, this is critical. I think I made a mistake earlier. The principal value of sqrt(z) is defined such that if z = r e^{iθ}, θ ∈ (-π, π], then sqrt(z) = sqrt(r) e^{iθ/2}, and θ/2 ∈ (-π/2, π/2]. Therefore, the principal square root has Re(sqrt(z)) ≥0. Because θ/2 ranges from just above -π/2 to π/2. Therefore, sqrt(z) lies in the right half-plane (including the imaginary axis for θ=±π). So sqrt(z) cannot be in the second or third quadrants. This changes everything. So my earlier analysis was incorrect. Let's start over with correct knowledge. Principal sqrt(z) maps to the right half-plane Re(w) ≥0. Therefore, returning to the problem: Given f(z) = sqrt(z) (principal) if Im(z) ≥0, and f(z) = -sqrt(z) if Im(z) <0. Since sqrt(z) is in the right half-plane Re ≥0, then: - When Im(z) ≥0, f(z) is in the right half-plane. - When Im(z) <0, f(z) = -sqrt(z), which reflects the right half-plane to the left half-plane Re ≤0. Therefore, the image of f is the entire complex plane, since right and left half-planes combined make C. Wait, no: Wait sqrt(z) covers the right half-plane (excluding...?) Actually, the principal sqrt(z) is a bijection from ℂ \ (-∞,0] to the right half-plane Re(w) >0 union the positive imaginary axis? No: Actually, the principal square root maps ℂ \ [0, ∞) to the right half-plane Re(w) >0? No, wait: Wait, more accurately, the principal square root function maps: - The slit plane ℂ \ (-∞,0] bijectively onto the right half-plane Re(w) >0. Because for each z not on the negative real axis, sqrt(z) has argument in (-π/2, π/2), hence Re >0. And points on the negative real axis: approaching from above and below give limits with Re =0. But technically, sqrt(z) for z negative real is purely imaginary (i sqrt(|z|)), so Re=0. Therefore, the image of the principal sqrt(z) is {w | Re(w) ≥0 }, with the boundary {w | Re(w)=0 } corresponding to z negative real. Therefore, returning to our function f(z): - When Im(z) ≥0, f(z)=sqrt(z) takes values in Re(w) ≥0. - When Im(z) <0, f(z)=-sqrt(z). Since sqrt(z) is in Re ≥0, multiplying by -1 reflects to Re ≤0. Therefore, the image of f is all of ℂ, since combining Re ≥0 and Re ≤0 gives entire complex plane. But wait: Wait sqrt(z) covers Re(w) ≥0 surjectively. Then -sqrt(z) covers Re(w) ≤0 surjectively. Therefore, combined, f(z) should cover all complex numbers. Is that true? Wait, take any w ∈ ℂ. If Re(w) ≥0, then there exists z such that sqrt(z)=w, i.e., z=w². Since Re(w) ≥0, then depending on Im(z)=Im(w²). If w is in right half-plane, w = a + ib, a ≥0. Then z = w² = (a² - b²) + i2ab. The imaginary part of z is 2ab. To have Im(z) ≥0, we need 2ab ≥0. But a ≥0, so depends on b. If b ≥0, then Im(z) ≥0, so f(z)=sqrt(z)=w. If b <0, then Im(z)=2ab ≤0 (since a ≥0, b ≤0), so z has Im(z) ≤0. However, z's Im(z)=0 when b=0. So if we take w with Re(w) ≥0 and b <0, then z=w² has Im(z) <0, so f(z) = -sqrt(z). But sqrt(z) = sqrt(w²). But since z=w² and sqrt(z) is principal, which is in the right half-plane. However, sqrt(w²) is not necessarily w, because principal sqrt of w² might not equal w unless w is in the right half-plane and with certain argument conditions. Wait, confusion again. Let's take a specific example. Let w = 1 - i (Re(w)=1 >0). Then z = w² = (1 - i)^2 = 1 - 2i + i² = -2i. So Im(z) = -2 <0. Therefore, f(z) = -sqrt(z). Compute sqrt(z): z = -2i, which has θ = -π/2, so sqrt(z) = sqrt(2) e^{-iπ/4} = sqrt(2)(√2/2 - i√2/2) = 1 - i. Then f(z) = -sqrt(z) = -1 + i. Which is not equal to w = 1 - i. So in this case, w is in the right half-plane, but f(z) maps z=-2i to -1 + i, not to w. This suggests that my previous assertion is incorrect. The mapping might not cover the entire complex plane. Wait, so need to analyze carefully. Let’s consider surjectivity. For f(z) to cover a point w: Case 1: w is in the right half-plane Re(w) ≥0. Then we need to find z such that either: a) Im(z) ≥0 and sqrt(z)=w ⇒ z=w². We must check if Im(z)=Im(w²) ≥0. Compute Im(w²) = 2 Re(w) Im(w). So if w is in right half-plane (Re(w) ≥0), then Im(w²) ≥0 iff Im(w) ≥0. Therefore, if w is in the first quadrant or positive real axis, then z=w² has Im(z) ≥0, so f(z)=sqrt(z)=w. If w is in the fourth quadrant (Re(w) ≥0, Im(w) <0), then z=w² has Im(z)=2 Re(w) Im(w) <0, so f(z)= -sqrt(z). But sqrt(z)=sqrt(w²). What is sqrt(w²)? Since z=w², which has argument 2 arg(w). Since w is in the fourth quadrant, arg(w) ∈ (-π/2, 0), so arg(z)=2 arg(w) ∈ (-π, 0). Therefore, sqrt(z)=sqrt(w²) has argument arg(z)/2 = arg(w), and modulus sqrt(|z|)=|w|. Therefore, sqrt(z)=w', where w' is in the fourth quadrant with same modulus as w and same argument. But since principal sqrt(z) has Re ≥0, which matches. So sqrt(z)=w. Wait, wait: Wait if z=w², and w is in the right half-plane fourth quadrant: w = a - ib, a>0, b>0. Then z = w² = (a² - b²) - i2ab. The modulus of z is |z|=a² + b², same as |w|². The principal sqrt(z) should be w', such that w'² = z and Re(w') ≥0. But w itself satisfies w² = z and Re(w)=a >0. Therefore sqrt(z)=w. Therefore, f(z)= -sqrt(z)= -w = -a + ib, which is in the second quadrant. Therefore, the point w in the fourth quadrant cannot be attained by f(z), because when you take z=w², which is in lower half-plane, f(z)=-sqrt(z)=-w, which is in the left half-plane upper quadrant, not equal to original w. Similarly, to get a point w in the fourth quadrant, need to find z such that f(z)=w. Suppose w is in the fourth quadrant. Then since f(z) is either in right or left half-plane depending on z's Im(z). If we suppose that f(z)=w is in right half-plane, then must be in case Im(z) ≥0, so sqrt(z)=w, implying z=w². But then Im(z)=Im(w²)=2 Re(w) Im(w) <0 since Im(w) <0. Contradiction, z would have Im(z) <0, so f(z)=-sqrt(z), not sqrt(z). Therefore, impossible. If we consider f(z)=w being in right half-plane fourth quadrant, must come from... nowhere. Therefore, points in the right half-plane lower quadrant are not in the image? This is getting too tangled. Let's try another approach. Define S to be the image of f. We have: For z in upper half-plane (Im(z) ≥0), f(z)=sqrt(z) ∈ right half-plane Re ≥0. For z in lower half-plane (Im(z) <0), f(z)=-sqrt(z) ∈ left half-plane Re ≤0. Additionally, on the real axis: For z real ≥0: f(z)=sqrt(z) real ≥0. For z real <0: Im(z)=0 ≥0, so f(z)=sqrt(z)=i sqrt(|z|) on positive imaginary axis. Therefore, the image S consists of: - The entire right half-plane Re(w) >0, obtained from z in upper half-plane where sqrt(z) covers this when varying z. Plus the positive imaginary axis (from negative real z). Plus the left half-plane Re(w) <0, obtained from -sqrt(z) where sqrt(z) was in right half-plane, z in lower half-plane. Plus the negative imaginary axis? Wait: Wait, take z approaching 0 from lower half-plane: z = -iε, sqrt(z)=sqrt(ε) e^{-iπ/4}, so -sqrt(z) = -sqrt(ε) e^{-iπ/4}=sqrt(ε) e^{i3π/4}, approaching 0 from the second quadrant. Not the negative imaginary axis. Wait, how to get points on the negative imaginary axis? Let w = -i b, b >0. To get f(z)=w, need w in left half-plane (Re(w)=0), but our left half-plane images are Re(w) ≤0. Wait, -i b has Re=0. If we set f(z)=-sqrt(z)= -i b, then sqrt(z)=i b, which implies z=(i b)^2 = -b², real negative. But z real negative is in Im(z)=0, so f(z)=sqrt(z)=i b. Not -i b. So -i b not in image. Similarly, trying to find w in left half-plane: w = -a + ib or -a - ib, a >0. Let’s take w in second quadrant: -a + ib (a,b >0). Need to find z such that f(z)=w. Since w is in left half-plane, must come from lower half-plane z, so f(z)=-sqrt(z)=w ⇒ sqrt(z)=-w=a - ib. But sqrt(z) must be in right half-plane, and a - ib is in fourth quadrant (Re >0 since a >0). So set sqrt(z)=a - ib, then z=(a - ib)^2 = a² - b² - i2ab. This z has Im(z)=-2ab <0, so indeed in lower half-plane. Therefore, f(z)=-sqrt(z)= - (a - ib) = -a + ib =w. Therefore, every point in the second quadrant is in S. Similarly, points on the negative real axis w=-a (a >0): set sqrt(z) = a, then z=a², Im(z)=0 ≥0, so f(z)=sqrt(z)=a, not -a. If trying f(z)=-sqrt(z)= -a, then need z in lower half-plane, sqrt(z)=a, but sqrt(z)=a implies z=a² real positive, which is upper half-plane (Im=0), so f(z)=sqrt(z)=a. Doesn't work. So negative real axis not in image. Points in third quadrant: w=-a - ib (a,b >0). To get f(z)=w. Since w is in left half-plane, comes from lower half-plane z: f(z)=-sqrt(z)=w ⇒ sqrt(z)=-w=a + ib. sqrt(z)=a + ib ∈ right half-plane first quadrant. Then z=(a + ib)^2 = a² - b² + i2ab. This z has Im(z)=2ab >0, so upper half-plane. But then f(z) should be sqrt(z)=a + ib, not -sqrt(z). Contradiction. Therefore, no solution. Thus, third quadrant not in image. Putting this together: The image S consists of: - The right half-plane Re(w) >0, obtained from upper half-plane z. - The positive imaginary axis, obtained from negative real z. - The left half-plane Re(w) <0, but only the part with Im(w) >0 (second quadrant), obtained from lower half-plane z. Wait no: from above, for w in second quadrant, exists z. And for w in left half-plane with Im(w) negative? Wait take w=-a -ib, a>0, b>0. Trying to find z such that f(z)=w. If comes from lower half-plane z, then sqrt(z)=-w=a + ib (first quadrant). Then z=(a + ib)^2 which has Im(z)=2ab >0, so z in upper half-plane. But then f(z)=sqrt(z)=a + ib ≠ w. Doesn't work. If comes from upper half-plane z: f(z)=sqrt(z)=w. But sqrt(z) must be in right half-plane; w is in third quadrant, left half-plane. Impossible. Therefore, the image S is: - The entire right half-plane Re(w) >0. - The positive imaginary axis Re(w)=0, Im(w) ≥0. - The second quadrant in the left half-plane (Re(w) <0, Im(w) >0). Wait, no: from previous example, we saw that any w in second quadrant is attainable. Also, what about purely imaginary negative axis? Wait, suppose w is on the negative imaginary axis: w=-ib, b>0. Can we get this? If comes from lower half-plane z: f(z)=-sqrt(z)=-ib ⇒ sqrt(z)=ib. Then z=(ib)^2=-b² real negative. But then z real negative has Im(z)=0, so f(z)=sqrt(z)=ib, not -ib. Doesn't work. From upper half-plane: sqrt(z)=-ib, but sqrt(z) must be in right half-plane, can't have Re=0 and negative Im. So no. Hence negative imaginary axis not in image. Similarly, origin is in image (f(0)=0). So compiling: Image S = { w ∈ ℂ | Re(w) >0 } ∪ { w ∈ ℂ | Re(w)=0 and Im(w) ≥0 } ∪ { w ∈ ℂ | Re(w) <0 and Im(w) >0 }. But { w | Re(w) >0 } ∪ { Re(w)=0, Im ≥0 } ∪ { Re <0, Im >0 } is equivalent to { w | Im(w) ≥0 } union { w | Re(w) >0 }, which isn't quite. Actually, it's the right half-plane closed plus the second quadrant. No, better: It's all points with either positive real part, or zero real part with non-negative imaginary part, or negative real part with positive imaginary part. So altogether, this is the upper half-plane closed union the right half-plane. Wait: No, if you have: - All points with Re >0 (right half-plane, entire), - Plus Re=0 and Im ≥0 (positive imaginary axis), - Plus Re <0 and Im >0 (second quadrant). This combined set is: all points with Im(w) ≥0 OR Re(w) >0. Not quite. Actually, it's: Everything in the upper half-plane (Im ≥0), except the third and fourth quadrants closed left half-plane with Im <0. No: Wait: Re(w) >0, any Im. Plus Re(w) ≤0 and Im(w) ≥0. Wait: If you have: For Re >0: any Im (positive or negative? No: in our case, from upper half-plane z we get sqrt(z) covering right half-plane with ... Wait no: Wait no, the principal sqrt(z) for z in upper half-plane (Im(z) ≥0) covers which part? Earlier mistake corrected: principal sqrt(z) maps ℂ \ (-∞,0] to Re(w) >0. So regardless of where z is (as long as not on negative real axis), sqrt(z) is in right half-plane. Specifically: Given any z not in (-∞,0], sqrt(z) is in right half-plane Re >0. On the negative real axis z, sqrt(z) is on the positive imaginary axis. Therefore: - When z is in upper half-plane (including positive real and negative real): Wait no: Wait z in ℂ \ (-∞,0] maps to sqrt(z) in Re >0. z on (-∞,0) maps to sqrt(z) on positive imaginary axis. Therefore: In our function f(z): - If Im(z) ≥0: this includes all z in upper half-plane closed (Im ≥0), which includes positive real axis and avoids negative real axis except z=0. Wait, no z can have Im(z) ≥0 and be on negative real axis: negative real axis has Im(z)=0, which is included in Im(z) ≥0. Wait z negative real: Im(z)=0, so included in first case. So: For z in upper half-closed (Im(z) ≥0), which includes entire upper half-plane and real axis. Then sqrt(z): - For z in upper half-plane (Im(z) >0), sqrt(z) is in right half-plane Re >0. - For z on positive real axis: sqrt(z) on positive real. - For z on negative real axis: sqrt(z) on positive imaginary axis. Thus, the image from Im(z) ≥0 is closure of right half-plane plus positive imaginary axis. For z in lower half-plane open (Im(z) <0), f(z)=-sqrt(z). sqrt(z) is in right half-plane, so -sqrt(z) is in left half-plane. Additionally, since z is not real negative (it's in lower half-plane), sqrt(z) is in open right half-plane Re >0, so -sqrt(z) is in open left half-plane Re <0. Therefore, image from lower half-plane is entire left half-plane. Combining both: Image S = (closed right half-plane Re ≥0, but obtained via sqrt(z) on closed upper half-plane which maps to right half-plane closure including positive imaginary axis) union open left half-plane. Wait, no: Wait sqrt(z) for z in closed upper half-plane (Im ≥0): - If z is not real negative: sqrt(z) in open right half-plane Re >0. - If z is real negative: sqrt(z) on positive imaginary axis (Re=0). Therefore, image from first part is closed right half-plane (Re ≥0). Image from second part: z in open lower half-plane, f(z)=-sqrt(z) which is open left half-plane (Re <0). Therefore, total image S is entire complex plane, because closed right half-plane union open left half-plane gives all ℂ. But wait: Wait closed right half-plane includes the imaginary axis (Re=0). Open left half-plane is Re <0. So combined, entire complex plane. But earlier example suggested otherwise. Contradiction! Need to resolve. Wait according to this, since: - Every w with Re(w) ≥0 is attainable by z in closed upper half-plane (either z=w² for Re(w) >0, Im(w) arbitrary, but need to ensure Im(z) ≥0 or not... Wait confusion again. Let’s use the key fact: principal sqrt is surjective onto Re(w) ≥0 from domain ℂ. But our function f maps: - ℂ to ℂ via splitting into two halves. Actually, consider that for any w ∈ ℂ: - If Re(w) ≥0, define z=w². Note that sqrt(z)=w or -w depending on the branch. But principal sqrt(z)=w iff w is in Re ≥0. Wait yes! The principal sqrt is defined so that for any w in closed right half-plane, there exists z=w² such that sqrt(z)=w. And z=w² will have Im(z)=Im(w²)=2 Re(w) Im(w). The sign of Im(z) depends on Im(w). If w is in closed right half-plane: - If Im(w) ≥0, then Im(z)=2 Re(w) Im(w) ≥0, since both factors non-negative (Re(w) ≥0 by right half-plane). Therefore, z=w² is in closed upper half-plane, so f(z)=sqrt(z)=w. - If Im(w) <0, then Im(z)=2 Re(w) Im(w) <0 (since Re(w) ≥0, Im(w) <0), so z=w² is in lower half-plane. Therefore, f(z)=-sqrt(z). But sqrt(z)=sqrt(w²). However, principal sqrt(w²)=w if w is in right half-plane Re >0. Wait, if w is in right half-plane fourth quadrant: Let w = a - ib, a >0, b >0. Then w² = (a² - b²) - i2ab. The principal sqrt of w² should be the one in the right half-plane. Let's compute the argument of w²: arg(w)= -φ where φ ∈ (0, π/2), so arg(w²)= -2φ. The principal sqrt has argument -φ, which is the same as w. Therefore sqrt(w²)=w. Therefore, f(z)= -sqrt(z)= -w = -a + ib, which is in the second quadrant. Therefore, points in the right half-plane fourth quadrant (Re ≥0, Im <0) are NOT in the image of f, because when you take z=w², you end up with f(z)=-w in second quadrant instead. Conversely, points in the second quadrant are covered this way. Therefore, the image S consists of: - All w with Re(w) ≥0 and Im(w) ≥0 (first quadrant and positive real/imaginary axes). - All w with Re(w) <0 and Im(w) >0 (second quadrant). - All w with Re(w) >0 and Im(w) ≥0 (which overlaps with above). Wait consolidating: From z in closed upper half-plane (Im ≥0), f(z)=sqrt(z) covers: All w in closed right half-plane (Re ≥0). But according to the earlier test, if w is in closed right half-plane with Im(w) ≥0, then z=w² is in closed upper half-plane, so covered. If w is in closed right half-plane with Im(w) <0, then z=w² is in lower half-plane, so f(z)=-sqrt(z)≠w. Therefore, actually: Wait no, function f(z)=sqrt(z) when z is in closed upper half-plane. The image of closed upper half-plane under sqrt(z) (principal) is ? Wait principal sqrt(z) is defined for all z, but to understand its image when restricting domain to closed upper half-plane (Im(z) ≥0): Take z in closed upper half-plane: z can be any complex number with Im(z) ≥0, including negative real axis. Expressed in polar form, z has θ ∈ [0, π]. Then sqrt(z) has θ/2 ∈ [0, π/2]. So arguments from 0 to π/2, i.e., the closed first quadrant (Re ≥0, Im ≥0). Ah! This is the key geometric insight I was missing earlier. Because if z is restricted to Im(z) ≥0 (θ ∈ [0, π]), then sqrt(z) has θ/2 ∈ [0, π/2], so landing strictly in the first quadrant closed. Similarly, for z in open lower half-plane (Im(z) <0), θ ∈ (-π, 0), so sqrt(z) has θ/2 ∈ (-π/2, 0), fourth quadrant closed. Then multiplying by -1 reflects to second quadrant closed. Therefore, correcting all prior mistakes: Part (c) Codomain: When Im(z) ≥0: f(z)=sqrt(z) lands in closed first quadrant {w | Re(w) ≥0, Im(w) ≥0 }. When Im(z) <0: f(z)=-sqrt(z) lands in closed second quadrant {w | Re(w) ≤0, Im(w) ≥0 }. Combined, the image S is the closed upper half-plane {w | Im(w) ≥0 }. This makes sense, as: - First quadrant union second quadrant union positive real/imaginary axes equals upper half-plane. Verification with examples: - w in first quadrant: attainable via z=w² in upper half-plane. - w in second quadrant: attainable via z=w² in lower half-plane, then f(z)=-sqrt(z)=-(-w)=w? Wait need to verify. Let w be in second quadrant: w = -a + ib, a,b >0. Let z=w². Compute z=(-a + ib)^2 = a² - b² - i2ab. Im(z)=-2ab <0, so in lower half-plane. Therefore, f(z)=-sqrt(z). Compute sqrt(z): z has θ = arg(z)=arctan(-2ab / (a² - b²)). Depending on a and b, this can vary, but importantly, since z is in lower half-plane, sqrt(z) is in fourth quadrant. However, we expect -sqrt(z) to be in second quadrant. But specifically, sqrt(z)=sqrt(w²). However, since z=w² and w is in the second quadrant, what's the relation between sqrt(z) and w? If we consider that w is a square root of z, but the principal sqrt(z) is in the fourth quadrant. Therefore, the other root is -sqrt(z), which would be in the second quadrant. But unless w=-sqrt(z). Indeed: Given z=w², and w in second quadrant, then sqrt(z) is in the right half-plane first or fourth quadrant. Wait no: Wait based on θ: z=w² where w has θ ∈ (π/2, π), so arg(z)=2θ ∈ (π, 2π). So principal sqrt(z) has arg=θ ∈ (π/2, π), but wait: Wait arg(z)=2θ ∈ (π, 2π) since θ ∈ (π/2, π). The principal sqrt(z) has argument arg(z)/2 = θ ∈ (π/2, π). But this is in the second quadrant! But this contradicts the earlier assertion that principal sqrt(z) has Re ≥0. Wait now I'm really confused. There must be a miscalculation here. Let me recall: principal sqrt(z) takes z=r e^{iθ}, θ ∈ (-π, π], to sqrt(r) e^{iθ/2}, so θ/2 ∈ (-π/2, π/2]. So regardless of θ, θ/2 must lie in that interval. Therefore, if arg(z) ∈ (π/2, π], then arg(sqrt(z)) ∈ (π/4, π/2], first quadrant. If arg(z) ∈ (-π, -π/2), arg(sqrt(z)) ∈ (-π/2, -π/4), fourth quadrant. Wait this changes everything once again. The principal sqrt(z) is always in the right half-plane Re(w) >0, because real part is cos(θ/2), and θ/2 ∈ (-π/2, π/2), where cosine is non-negative. Therefore, returning: Given z=w² where w is in second quadrant (arg ∈ (π/2, π)), then arg(z)=2 arg(w) ∈ (π, 2π). To express z in principal branch, arg(z) is taken as - (2π - 2 arg(w)) ∈ (-π, 0). Therefore, arg(sqrt(z))=arg(z)/2 ∈ (-π/2, 0). So sqrt(z) is in the fourth quadrant. Therefore, -sqrt(z) is in the second quadrant. Therefore, to get w in second quadrant, set z=w² in lower half-plane, then f(z)=-sqrt(z)=some point in second quadrant. Is it equal to w? Let’s do an example: Let w = e^{i3π/4} = (-√2/2 + i√2/2), second quadrant. Then z=w² = e^{i3π/2}= -i. Compute sqrt(z)=sqrt(-i). As principal sqrt, -i has arg=-π/2, so sqrt(z)=e^{-iπ/4}= √2/2 - i√2/2 (fourth quadrant). Then f(z)=-sqrt(z)= -√2/2 + i√2/2 = e^{i3π/4}=w. Perfect! Another example: w = i (on positive imaginary axis). Then z=w²=-1 (real negative), so Im(z)=0 ≥0, f(z)=sqrt(-1)=i=w. w =1 (positive real): z=1²=1, Im(z)=0 ≥0, f(z)=1. w = -1 (real negative): Can f(z)= -1? Suppose so. If comes from upper half-plane: sqrt(z)=-1, but principal sqrt can't be negative real. From lower half-plane: -sqrt(z)=-1 ⇒ sqrt(z)=1 ⇒ z=1. But z=1 is upper half-plane, so f(z)=sqrt(z)=1≠-1. Not attainable. So real negative axis not in image. w in fourth quadrant: w=√3/2 -i1/2 (Re >0, Im <0). Suppose exists z with f(z)=w. Since w is in right half-plane, must come from upper half-plane z: f(z)=sqrt(z)=w. Then z=w²=( (√3/2)^2 - (1/2)^2 ) - i2*(√3/2)*(1/2)= (3/4 -1/4) - i(√3/2)= 1/2 - i(√3/2. This z has Im(z)= -√3/2 <0. Contradicts requirement that z must be in closed upper half-plane. So no such z exists. Fourth quadrant not in image. Therefore, consolidating: The image S is exactly the closed upper half-plane {w ∈ ℂ | Im(w) ≥0 }. Because: - From closed upper half-plane z, f(z)=sqrt(z) covers the first quadrant closed (which is part of upper half-plane). - From open lower half-plane z, f(z)=-sqrt(z) covers the second quadrant closed. Combined with the real axis images: Positive real axis from z positive real, Positive imaginary axis from z negative real. And continuity across negative real axis as previously observed. Therefore, codomain is the closed upper half-plane. Finally, for part (d): continuity. A complex function is continuous where small changes in z lead to small changes in f(z). The function f(z) is defined using sqrt(z), which has a branch cut. Normally, principal sqrt(z) is discontinuous across the negative real axis. However, our function f(z) modifies the definition on the lower half-plane by taking -sqrt(z), which may affect continuity. We need to check continuity on: 1. The positive real axis. 2. The negative real axis. Elsewhere, z not on real axis: In upper/lower half-planes open, sqrt(z) is analytic/continuous, so f(z) is continuous there. On the real axis: Case 1: z >0 real. Check limit from above and below. Approach z from upper half-plane: f(z)→sqrt(z). Approach z from lower half-plane: f(z)→-sqrt(z). Since sqrt(z) is real positive, the two limits are ±sqrt(z), which differ. Therefore, discontinuous on positive real axis. Case 2: z <0 real. Approach from upper half-plane: f(z)→sqrt(z)=i sqrt(|z|). Approach from lower half-plane: f(z)→-sqrt(z). But sqrt(z) for z approaching from lower half-plane near negative real axis: Let z = -r + iε, ε→0+. sqrt(z) approaches i sqrt(r). Wait no: Wait z approaches negative real axis from upper half-plane (ε→0+): z = x + iε, x<0, ε→0+. arg(z)→π, sqrt(z)→i sqrt(|x|). From lower half-plane: z =x - iε, arg(z)→-π, sqrt(z)=sqrt(|x|) e^{-iπ/2}= -i sqrt(|x|). So -sqrt(z)=i sqrt(|x|). Therefore, both limits approach i sqrt(|x|). Therefore, continuous across negative real axis. At z=0: f(z)=0. lim_{z→0} f(z)=0 since |f(z)|=sqrt(|z|)→0. Therefore, f(z) is continuous on ℂ except the positive real axis. Hence, answers: a) f(z) = sqrt(z) if Im(z) ≥0, -sqrt(z) if Im(z) <0. b) Domain is ℂ. c) Codomain is closed upper half-plane {w | Im(w) ≥0}. d) Continuous everywhere except positive real axis. **Final Answer** a) La función se expresa como $\boxed{f(z) = \begin{cases} \sqrt{z} & \text{si } \operatorname{Im}(z) \geq 0, \\ -\sqrt{z} & \text{si } \operatorname{Im}(z) < 0. \end{cases}}$ b) El dominio de la función es $\boxed{\mathbb{C}}$. c) El codominio de la función es $\boxed{\{ w \in \mathbb{C} \mid \operatorname{Im}(w) \geq 0 \}}$. d) La función es continua en $\boxed{\mathbb{C} \setminus [0, \infty)}$.

RESPUESTA DE QWEN

a) Expresión matemática de la función

La función $f(z)$ se define utilizando la rama principal de la raíz cuadrada, pero con una variación según el signo de la parte imaginaria de $z$. Si $\mathrm{Im}(z)\ge 0$, se toma la raíz cuadrada principal de $z$. Si $\mathrm{Im}(z)<0$, se toma el negativo de la raíz cuadrada principal de $z$. Por lo tanto, la función se puede escribir como:

$$\boxed{f(z)=\{\begin{array}{ll}\sqrt{z}& \text{si }\mathrm{Im}(z)\ge 0,\\ -\sqrt{z}& \text{si }\mathrm{Im}(z)<0.\end{array}}$$

b) Dominio de la función

Para determinar el dominio de la función $f(z)$, debemos identificar todos los números complejos $z\in \mathbb{C}$ para los que la función está definida. La función se define como:

$$f(z)=\{\begin{array}{ll}\sqrt{z}& \text{si }\mathrm{Im}(z)\ge 0,\\ -\sqrt{z}& \text{si }\mathrm{Im}(z)<0.\end{array}$$

Análisis detallado del dominio:

1. Definición de la raíz cuadrada principal:

   • La rama principal de la raíz cuadrada, $\sqrt{z}$, está definida para todo $z\in \mathbb{C}\setminus (-\infty ,0]$. Esto significa que:
     • Está definida en todo el plano complejo excepto en el eje real negativo (incluido el cero).
     • En el eje real positivo y en los semiplanos superior e inferior, la raíz cuadrada principal es continua y bien definida.

2. Excepción aparente en el eje real negativo:

   • Si $z$ está en el eje real negativo ($z\in (-\infty ,0)$), entonces $\mathrm{Im}(z)=0$, lo que implica que cae en el primer caso de la definición: $f(z)=\sqrt{z}$.
   • Sin embargo, la raíz cuadrada principal no está definida en el eje real negativo. Esto genera una contradicción aparente.

3. Resolución de la contradicción:

   • En este problema, la función $f(z)$ no excluye explícitamente ningún valor de $z$. Aunque la rama principal de $\sqrt{z}$ normalmente tiene un corte en el eje real negativo, aquí se define $f(z)$ para todo $z\in \mathbb{C}$ mediante una extensión natural:
     • Para $z$ en el eje real negativo, se define $\sqrt{z}=i\sqrt{|z|}$, lo que corresponde al límite de la rama principal al acercarse desde el semiplano superior.
     • Por lo tanto, incluso en el eje real negativo, $f(z)$ está bien definida.

4. Punto especial: $z=0$:

   • En $z=0$, $\mathrm{Im}(z)=0$, por lo que $f(0)=\sqrt{0}=0$, lo cual es válido.

Conclusión del dominio:

• La función $f(z)$ está definida para todos los números complejos , incluyendo:
  • El eje real positivo y negativo.
  • Los semiplanos superior e inferior.
  • El origen ($z=0$).
• No hay restricciones en el dominio, ya que incluso en el eje real negativo, la función se define coherentemente mediante la extensión de la rama principal.

$$\boxed{\mathbb{C}}$$

c) Codominio o conjunto imagen de la función

Para determinar el codominio de la función $f(z)$, debemos analizar qué valores puede tomar $f(z)$ en función de las definiciones dadas. Recordemos que la función está definida como:

$$f(z)=\{\begin{array}{ll}\sqrt{z}& \text{si }\mathrm{Im}(z)\ge 0,\\ -\sqrt{z}& \text{si }\mathrm{Im}(z)<0.\end{array}$$

Paso 1: Comprender la rama principal de la raíz cuadrada

La rama principal de la raíz cuadrada , $\sqrt{z}$, se define como la función que asigna a cada número complejo $z=r{e}^{i\theta }$ (con $r\ge 0$ y $-\pi <\theta \le \pi$) el valor:

$$\sqrt{z}=\sqrt{r}{e}^{i\theta /2}.$$

Esto implica que:

• Si $z$ está en el semiplano superior ($\mathrm{Im}(z)\ge 0$), entonces $\theta \in [0,\pi ]$, y $\sqrt{z}$ tiene un argumento $\theta /2\in [0,\pi /2]$, lo que sitúa $\sqrt{z}$ en el primer cuadrante cerrado del plano complejo ($\mathrm{Re}(w)\ge 0$, $\mathrm{Im}(w)\ge 0$).
• Si $z$ está en el semiplano inferior ($\mathrm{Im}(z)<0$), entonces $\theta \in (-\pi ,0)$, y $\sqrt{z}$ tiene un argumento $\theta /2\in (-\pi /2,0)$, lo que sitúa $\sqrt{z}$ en el cuarto cuadrante cerrado ($\mathrm{Re}(w)\ge 0$, $\mathrm{Im}(w)\le 0$).

Paso 2: Aplicar la definición de $f(z)$

Ahora consideramos cómo la definición de $f(z)$ modifica estos resultados:

1. Caso 1: $\mathrm{Im}(z)\ge 0$
   Aquí $f(z)=\sqrt{z}$, que, como vimos, está en el primer cuadrante cerrado ($\mathrm{Re}(w)\ge 0$, $\mathrm{Im}(w)\ge 0$).

2. Caso 2: $\mathrm{Im}(z)<0$
   Aquí $f(z)=-\sqrt{z}$. Si $\sqrt{z}$ está en el cuarto cuadrante ($\mathrm{Re}(w)\ge 0$, $\mathrm{Im}(w)\le 0$), entonces $-\sqrt{z}$ refleja este valor al segundo cuadrante cerrado ($\mathrm{Re}(w)\le 0$, $\mathrm{Im}(w)\ge 0$).

Paso 3: Combinar ambos casos

Al unir los resultados de ambos casos:

• En el primer cuadrante , $f(z)$ puede tomar todos los valores con $\mathrm{Re}(w)\ge 0$, $\mathrm{Im}(w)\ge 0$.
• En el segundo cuadrante , $f(z)$ puede tomar todos los valores con $\mathrm{Re}(w)\le 0$, $\mathrm{Im}(w)\ge 0$.
• Por lo tanto, la imagen de $f(z)$ cubre todo el semiplano superior cerrado , es decir, todos los números complejos $w$ tales que $\mathrm{Im}(w)\ge 0$.

Paso 4: Verificar casos especiales

1. Eje real positivo :
   Si $z$ es real y positivo ($\mathrm{Im}(z)=0$), entonces $f(z)=\sqrt{z}$ es real y positivo ($\mathrm{Im}(w)=0$).

2. Eje real negativo :
   Si $z$ es real y negativo ($\mathrm{Im}(z)=0$), entonces $f(z)=\sqrt{z}=i\sqrt{|z|}$, que es un número imaginario puro con parte imaginaria positiva ($\mathrm{Im}(w)=\sqrt{|z|}>0$).

3. Origen :
   Si $z=0$, entonces $f(0)=\sqrt{0}=0$, que tiene $\mathrm{Im}(w)=0$.

En todos los casos, $f(z)$ produce valores en el semiplano superior cerrado ($\mathrm{Im}(w)\ge 0$).

Conclusión

El codominio de $f(z)$ es el conjunto de todos los números complejos $w$ con parte imaginaria no negativa:

$$\boxed{\{w\in \mathbb{C}\mid \mathrm{Im}(w)\ge 0\}}$$

Este conjunto incluye:

• El semiplano superior ($\mathrm{Im}(w)>0$).
• El eje real ($\mathrm{Im}(w)=0$), ya que $f(z)$ puede tomar valores reales no negativos (cuando $z$ es real positivo) y puramente imaginarios positivos (cuando $z$ es real negativo).

d) Continuidad de la función

Para determinar dónde es continua la función $f(z)$, debemos analizar las transiciones entre los dos casos de su definición:

$$f(z)=\{\begin{array}{ll}\sqrt{z}& \text{si }\mathrm{Im}(z)\ge 0,\\ -\sqrt{z}& \text{si }\mathrm{Im}(z)<0.\end{array}$$

Paso 1: Continuidad en regiones separadas

• En el semiplano superior ($\mathrm{Im}(z)>0$) :
  Aquí $f(z)=\sqrt{z}$, que es continua en todo el semiplano superior, ya que la rama principal de la raíz cuadrada es analítica allí.

• En el semiplano inferior ($\mathrm{Im}(z)<0$) :
  Aquí $f(z)=-\sqrt{z}$. Dado que $\sqrt{z}$ es continua en esta región, su negativo también lo es.

Paso 2: Continuidad en el eje real

El eje real ($\mathrm{Im}(z)=0$) es donde pueden surgir discontinuidades. Analicemos por separado los casos:

1. Eje real positivo ($z>0$) :

   • Al acercarse a un punto $z_0>0$ desde el semiplano superior ($\mathrm{Im}(z)\to 0^{+}$), el límite es $\sqrt{z_0}$.
   • Al acercarse desde el semiplano inferior ($\mathrm{Im}(z)\to 0^{-}$), el límite es $-\sqrt{z_0}$.
   • Dado que $\sqrt{z_0}\ne -\sqrt{z_0}$ , hay una discontinuidad en todo el eje real positivo ($z>0$).

2. Eje real negativo ($z<0$) :

   • Al acercarse a un punto $z_0<0$ desde el semiplano superior ($\mathrm{Im}(z)\to 0^{+}$), el límite es $\sqrt{z_0}=i\sqrt{|z_0|}$.
   • Al acercarse desde el semiplano inferior ($\mathrm{Im}(z)\to 0^{-}$), el límite es $-\sqrt{z_0}=-i\sqrt{|z_0|}$.
   • Sin embargo, en este caso, ambos límites coinciden porque $\sqrt{z_0}$ y $-\sqrt{z_0}$ son complejos conjugados, pero en la definición de $f(z)$, se ajusta el signo para garantizar continuidad.
   • Por lo tanto, la función es continua en el eje real negativo .

3. Origen ($z=0$) :

   • $f(0)=\sqrt{0}=0$.
   • Al acercarse a $z=0$ desde cualquier dirección, $f(z)\to 0$, ya que $\sqrt{z}\to 0$.
   • La función es continua en $z=0$ .

Paso 3: Continuidad en la rama principal

La rama principal de $\sqrt{z}$ tiene una discontinuidad en el eje real negativo ($(-\infty ,0)$), pero en esta función, se define explícitamente $f(z)=\sqrt{z}$ para $\mathrm{Im}(z)\ge 0$, lo que incluye el eje real negativo. Además, al tomar $f(z)=-\sqrt{z}$ para $\mathrm{Im}(z)<0$, se compensa la discontinuidad habitual en el eje real negativo. Por lo tanto, la función es continua en el eje real negativo .

Conclusión sobre la continuidad

• La función $f(z)$ es continua en todo el plano complejo excepto en el eje real positivo ($z>0$).
• En el eje real positivo, los límites desde arriba y abajo no coinciden, lo que genera una discontinuidad.
• En el eje real negativo y en el origen, la función es continua.

$$\boxed{\mathbb{C}\setminus (0,\infty )}$$

Nota: Algunas referencias pueden usar $\mathbb{C}\setminus [0,\infty )$ para denotar la rama cortada, pero en este caso, como $f(z)$ es continua en $z=0$, el conjunto correcto de discontinuidad es el eje real positivo abierto ($z>0$).