11-02-2025

Problem: Find the real and imaginary part of the following functions $f: \mathbb{C} \to \mathbb{C} $.
  1. $f(z) = \sin z$
  2. $f(z) = z^2 \cos z$
  3. $f(z) = z^3 + 5z - 1$
  4. $f(z) = \frac{z^2 + 1}{z - 1}$
  5. $f(z) = \vert z \vert \bar{z}$
Solution: A general method to find the real and imaginary part of the complex function is to write $z = x + \iota y$ and then express the function in terms of $x$ and $y$. Separate the real and imaginary parts.

  1. For \( f(z) = \sin z \): \begin{align*} f(z) & = \frac{e^{\iota z} - e^{-\iota z}}{2\iota } \\[1ex] & = \frac{e^{\iota x - y} - e^{-\iota x + y}}{2\iota } \\[1ex] & = \frac{e^{\iota x} e^{-y} - e^{-\iota x} e^{y}}{2\iota } \\[1ex] & = \frac{e^{-y} \left( \cos x + \iota \sin x \right) - e^y \left( \cos x - \iota \sin x \right) }{2 \iota }\\ & = \frac{e^y + e^{-y} }{2} \sin x + \iota \frac{e^y - e^{-y} }{2 } \cos x. \end{align*} Therefore,
    \begin{align*} \text{Re} (f(z)) & = \frac{e^y + e^{-y} }{2} \sin x = \cosh y \sin x, \\[1ex] \text{Im} (f(z)) & = \frac{e^y - e^{-y} }{2 } \cos x = \sinh y \cos x. \end{align*}

  2. For the function \( f(z) = z^2 \cos z \): \[ z^2 = (x + \iota y)^2 = x^2 - y^2 + 2\iota xy, \] and by the same method or using the trigonometric identities we have \begin{align*} \cos z & = \cos (x + \iota y) \\ & = \cos x \cos \iota y - \sin x \sin \iota y. \end{align*} Since $x$ and $y$ are real, we have \begin{align*} & \cos(iy)=\frac{\exp(i^2y)+\exp(-i^2y)}{2}=\frac{\exp y+\exp(-y)}{2}=\cosh y \\ & \sin(iy)=\frac{\exp(i^2y)-\exp(-i^2y)}{2i}=-\frac{\exp y-\exp(-y)}{2i}=i\sinh y. \end{align*} Thus, \[ \cos z = \cos x \cosh y - \iota \sin x \sinh y. \] Therefore, \[ f(z) = (x^2 - y^2 + 2\iota xy)(\cos x \cosh y - \iota \sin x \sinh y). \] Separating the real and imaginary parts, we get:
    \begin{align*} \text{Re}(f(z)) & = (x^2 - y^2) \cos x \cosh y + 2xy \sin x \sinh y,\\ \text{Im}(f(z)) & = 2xy \cos x \sinh y - (x^2 - y^2) \sin x \cosh y. \end{align*}

  3. For \( f(z) = z^3 + 5z - 1 \): \begin{align*} z^3 & = (x + \iota y)^3 \\ & = x^3 + 3x^2 \iota y + 3x(\iota y)^2 + (\iota y)^3 \\ & = x^3 - 3xy^2 + \iota (3x^2 y - y^3). \end{align*} Therefore, \begin{align*} f(z) & = x^3 - 3xy^2 + \iota (3x^2 y - y^3) + 5 x + \iota 5y - 1 \\ & = x^3 - 3xy^2 + 5x - 1 + \iota (3x^2 y - y^3 + 5y). \end{align*} Thus,
    \begin{align*} \text{Re} (f(z)) & = x^3 - 3xy^2 + 5x - 1, \\ \text{Im} (f(z)) & = 3x^2 y - y^3 + 5y. \end{align*}

  4. For \( f(z) = \frac{z^2 + 1}{z - 1} \): \begin{align*} \frac{z^2 + 1}{z - 1} & = \frac{(x + \iota y)^2 + 1}{x + \iota y - 1} \\ & = \frac{(x^2 - y^2 + 1) + 2\iota xy}{(x - 1) + \iota y}. \end{align*} Multiplying the numerator and denominator by the conjugate of the denominator, we get: \begin{align*} f(z) & = \frac{((x^2 - y^2 + 1) + 2\iota xy)((x - 1) - \iota y)}{(x - 1)^2 + y^2}. \end{align*} Separating the real and imaginary parts, we get:
    \begin{align*} \text{Re}(f(z)) & = \frac{(x^2 - y^2 + 1)(x - 1) + 2xy^2}{(x - 1)^2 + y^2} \\ \text{Im}(f(z)) & = \frac{2xy(x - 1) - (x^2 - y^2 + 1)y}{(x - 1)^2 + y^2}. \end{align*}

  5. For \( f(z) = \vert z \vert \bar{z} \): \[ \vert z \vert = \sqrt{x^2 + y^2}, \text{ and } \bar{z} = x - \iota y. \] Therefore, \[ f(z) = \sqrt{x^2 + y^2}(x - \iota y) = x\sqrt{x^2 + y^2} - \iota y\sqrt{x^2 + y^2}. \] Thus,
    \begin{align*} \text{Re} (f(z)) & = x\sqrt{x^2 + y^2}, \\ \text{Im} (f(z)) & = -y\sqrt{x^2 + y^2}. \end{align*}
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