What is the image of $D=\{z:0<\operatorname{Re}z<\pi\}\setminus\{\pi/2\}$ under $f(z)=\tan z$?

Question

What would be the image of the domain $D = \{z:0<\operatorname{Re}z<\pi\} \setminus \{\pi/2\}$ under $f(z) = \tan z$?

I havn't met with tan(z) transformation so I don't really know how to start..

Answer 1

Start by writing $\tan z$ in terms of the functions you already met. $$\tan z = \frac{\sin z}{\cos z} = \frac{1}{i}\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{iz}} = \frac{1}{i}\frac{e^{2iz}-1}{e^{2iz}+1} \tag1$$ Therefore, the function is a composition of transformations $z\mapsto 2iz$, $z\mapsto e^z$, $z\mapsto (z-1)/(z+1)$, and $z\mapsto z/i$. Follow them step by step, recording what happens to the domain.

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Addition concerning $(z-1)/(z+1)$: fractional linear transformations are bijections on $\mathbb C\cup \{0\}$, and they preserve lines and circles (although lines can get mapped to circles and vice versa).

Of course, the image of $-1$ is just one point that you get by plugging $-1$ into $(z-1)/(z+1)$.

You can find the image of $[0,\infty]$ by picking three points from this half-line and marking their images, then drawing a line or a circle through them. Since $0\mapsto -1$, $1\mapsto 0$, $\infty\mapsto 1$, the image of $[0,\infty]$ is the segment $[-1,1]$.