# Finding the image of $D=\{z|\,-\frac{\pi}{2}<Re(z)<\frac{\pi}{2}\}$ under $f(z)=e^{iz}$

I am trying to solve the following question (and to generally understand how to solve these type of questions, i.e finding the $f(D)$ for some given $f,D$).

Let $D=\{z\in\mathbb{C}|\,-\frac{\pi}{2}<Re(z)<\frac{\pi}{2}\}$and $f(z)=e^{iz}$. Find $f(D)$

Where by find the question means to geometrically understand (using calculations, not neccaseraly geometric arguments ) where does $D$ maps to under $f$.

I don't really know where to start, I tried to see what happens for $Re(z)=\pm\frac{\pi}{2}$:

For $D_{1}=\{z\in\mathbb{C}|\, Re(z)=\frac{\pi}{2}\}$ I got that $f(D_{1})=\{yi|\, y>0\}$. Similarly I got $\{yi|\, y<0\}$ for $Re(z)=-\frac{\pi}{2}$.

I'm not even sure what does these calculations help me with, and if I should of done them, but I don't really know how to approach the question.

Can someone please help me understand how to solve this question, and what is the main idea I should follow on such questions ?

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Surely you can determine $f(D_\theta)$, where $D_\theta=\{z\in\mathbb C\mid\Re(z)=\theta\}$. – Did Feb 9 '13 at 20:35
@Did - Thanks! I got the idea here. So an approach to these type of questions is to think of a way to write $D$ as a disjoint union of lines ? – Belgi Feb 9 '13 at 20:39
I would not generalize too much the applicability of this principle but yes, here this approach simplifies things. // Are you going to write down a solution? – Did Feb 9 '13 at 20:44
Too late... Too bad. – Did Feb 9 '13 at 20:45

Your approach does tell us something. Given a conformal map (analytic function with everywhere non-$0$ derivative) $g:\Bbb C\to\Bbb C$ and a region (open connected set) $\Omega\subseteq\Bbb C$, if we allow $z\in\Omega$ to tend toward the boundary of $\Omega$, then $g(z)$ will tend toward the boundary of $g(\Omega)$. Since $f(z)=e^{iz}$ is a conformal map (check that), then your work shows us that the positive and negative imaginary axes will form part (in fact, almost all) of the boundary of $f(D)$. Moreover, if we take $z\in D$ and let $\text{Im}(z)$ tend toward $+\infty$, we see that $f(z)$ tends to $0$ (check that), so the origin is also part of the boundary of $f(D)$. This suggests that $f(D)$ will be a half-plane determined by the imaginary axis. Let's confirm that, and determine which one.

Suppose $z=x+iy$, where $x,y$ real and $|x|<\frac\pi2$--that is, suppose $z\in D$. Then \begin{align}f(z) &= e^{iz}\\ &= e^{ix-y}\\ &= e^{-y}e^{ix}\\ &= e^{-y}\bigl(\cos x+i\sin x\bigr).\end{align} Since there were no restrictions placed on $y$, then $e^{-y}$ may be any positive number. Since $|x|<\frac\pi2$, then $\sin x$ could be any number strictly between $-1$ and $1$, so since $e^{-y}$ could be any positive number, then $e^{-y}\sin x$--the imaginary part of $f(z)$--could be any real number. On the other hand, we must have $0<\cos x\leq 1$, and so the real part of $f(z)$ is necessarily positive. Hence, $$f(D)\subseteq\{w:\text{Re}(w)>0\}.$$

Now let's suppose $w=u+iv$, where $u,v$ real and $u>0$--that is, $w\in\{w:\text{Re}(w)>0\}.$ Since $u,v$ real and $u\neq 0$, then $|w|=\sqrt{u^2+v^2}>0$. Putting $y=-\ln\sqrt{u^2+v^2}$, we have that $y$ is real and $$e^{-y}=e^{\ln\sqrt{u^2+v^2}}=\sqrt{u^2+v^2}=|w|.$$ Putting $x=\arctan\frac{v}{u}$, we have $|x|<\frac\pi2$, that $\cos x=\frac{u}{\sqrt{u^2+v^2}},$ and that $\sin x=\frac{v}{\sqrt{u^2+v^2}}$. Thus, putting $z=x+iy$, we have $z\in D$ and $$f(z)=e^{-y}\bigl(\cos x+i\sin x\bigr)=\sqrt{u^2+v^2}\left(\frac{u}{\sqrt{u^2+v^2}}+\frac{iv}{\sqrt{u^2+v^2}}\right)=u+iv=w,$$ so $w\in f(D)$. Therefore, $$f(D)\supseteq\{w:\text{Re}(w)>0\},$$ so $$f(D)=\{w:\text{Re}(w)>0\}$$ by double inclusion.

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I find polar coordatines make these types of questions more clear. Let $z=a+bi$ be some complex number in the strip $D$. Then carefully exponentiating, $$f(z)=e^{i(a+bi)}=e^{-b}e^{ia}.$$ Since $b$ is unrestricted in $D$, $e^{-b}$ ranges over all positive real values. Since $-\pi/2<a<\pi/2$, the image $f(D)$ is the points with positive radius and and polar angle $-\pi/2<a<\pi/2$. This is of course just the right half plane.

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$$f(z) = e^{i z} = e^{i x} e^{-y} = e^{-y} \cos{x} + i e^{-y} \sin{x}$$
where $x=\Re{z}$ and $y=\Im{z}$. In this case, $x \in \{-\pi/2,\pi/2\}$ and $y \in (-\infty,\infty)$, so that $\Re{f(z)} \in (0,\infty)$ and $\Im{f(z)} \in (-\infty,\infty)$. Therefore, $f$ maps the given region into the right-half plane.
Thanks for the answer! How did you determine $Re(f(z)),Im(f(z))$ ? – Belgi Feb 9 '13 at 20:49
I thought I was pretty explicit about this in the post. I used the fact that $e^{i x} = \cos{x} + i \sin{x}$. Because e^{-y} is real, then $\Re{f(z)} = e^{-y} \cos{x}$ and $\Im{f(z)} = e^{-y} \sin{x}$. – Ron Gordon Feb 9 '13 at 20:52
Yes, I'm sorry if I wasn't clear - I meant how did you get that $Re(f(z))\in[0,\infty),Im(f(z))\in(-\infty,\infty)$. I understand that now, but I am having doubts that generally the answer is the Cartesian product of those...it seems important to know thing like what are the possible values of the imaginary part when the real part is fixed – Belgi Feb 9 '13 at 20:56
The region $D$ states that $x \in (-\pi/2,\pi/2)$. This means that $\cos{x} \in (0,1)$ and $\sin{x} \in (-1,1)$. Meanwhile, because there is no restriction on $y$, then $y \in (-\infty,\infty)$ and therefore $e^{-y} \in (0,\infty)$. – Ron Gordon Feb 9 '13 at 20:59
True, but I don't think that if $Re(f(D))=A$ and $Im(f(D))=B$ then $f(D)=A\times B$ which is seems to be what you used. Please correct me if I'm wrong, I am saying that it is important not only to know what are the values that the real and imaginary parts can take - but also for a fixed real part what is the range of the imaginary part – Belgi Feb 9 '13 at 21:03