Given is the function $f : \mathbb{C} \to \mathbb{C}$ defined by the expression $f(z) = \sinh z$.
(a) Prove that for all $z, w \in \mathbb{C}$, the following holds: $\sinh(z + w) = \sinh z \cosh w + \cosh z \sinh w$. Then calculate the real and imaginary parts of the function $f$.
(b) Determine where the function $f$ maps the line segment $D_a = \{ z \in \mathbb{C} \mid \operatorname{Re} z = a, \operatorname{Im} z \in [-\frac{\pi}{2}, \frac{\pi}{2}] \}$, where $a \in \mathbb{R}$ is a constant. Where does the region $[0, \infty) \times [-\frac{\pi}{2}, \frac{\pi}{2}]$ map to? Sketches are mandatory.
Attempt: (a) To prove the identity $\sinh(z + w) = \sinh z \cosh w + \cosh z \sinh w$ for $z, w \in \mathbb{C}$, I use the definitions $\sinh z = \frac{e^z – e^{-z}}{2}$ and $\cosh z = \frac{e^z + e^{-z}}{2}$. Are then the real and imaginary parts of $\sinh(x + iy)$ are $\sinh x \cos y$ and $\cosh x \sin y$ respectively?
(b) I think the line segment $D_a = \{ z \in \mathbb{C} \mid \operatorname{Re} z = a, \operatorname{Im} z \in [-\frac{\pi}{2}, \frac{\pi}{2}] \}$ maps to an ellipse in the complex plane with semi-major axis $\cosh a$ along the imaginary axis and semi-minor axis $\sinh a$ along the real axis. But I really don't know how to justify if this is even true.
Best Answer
For $a)$ note that for any $z \in \mathbb{C}$ we have that $\cosh(iz) = \cos(z)$ and $\sinh(iz) = i \sin(z)$.
So, for $z = x+iy \in \mathbb{C}$ we have that \begin{align*}\sinh(z) = \sinh(x+iy) &= \sinh(x) \cosh(iy) + \cosh(x) \sinh(iy) \\ &= \sinh(x) \cos(y) + \cosh(x) (i \sin(y)) \\&= \sinh(x) \cos(y) + i \cosh(x) \sin(y). \end{align*} So, $\Re(\sinh(x+iy)) = \sinh(x) \cos(y)$ and $\Im(\sinh(x+iy)) = \cosh(x) \sin(y)$.
For $b)$: Let $a \in \mathbb{R}$ fixed and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ arbitrary. If we set $x:= \sinh(a)\cos(\theta)$ and $y:=\cosh(a)\sin(\theta)$, then we have that $\frac{x^2}{(\sinh(a))^2} + \frac{y^2}{(\cosh(a))^2} = \cos^2(\theta) + \sin^2(\theta) = 1$. So, $(x,y)$ is a point on the ellipse $\mathcal{E_a}$ of semi-major axis $\sinh(a)$ and semi-minor axis $\cosh(a)$. Because $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we get that $f(D_a) = \mathcal{E_a}$.
And, we also have $f\left([0,\infty) \times [-\frac{\pi}{2}, \frac{\pi}{2}]\right) = f(0) \cup f(\bigcup_{a \in (0,\infty)} D_a) = f(0) \cup (\bigcup_{a \in (0,\infty)}f(D_a)) = \{0\} \cup \bigcup_{a \in [0, \infty)}\mathcal{E_a} = \\ = \{0\} \cup (\mathbb{C}\setminus\{0\}) = \mathbb{C}$