Problem: Find the real and imaginary part of the following functions $f: \mathbb{C} \to \mathbb{C} $.
- $f(z) = \sin z$
- $f(z) = z^2 \cos z$
- $f(z) = z^3 + 5z - 1$
- $f(z) = \frac{z^2 + 1}{z - 1}$
- $f(z) = \vert z \vert \bar{z}$
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.
-
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*}
-
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*}
-
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*}
-
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*}
-
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*}