The graph of $e^z$ for $|z|=1$

Question

When $|z|=1$(unit circle in complex plane), what is the graph of $e^z$? I know that this transformation makes a band into a angle area, but I have no idea about this .

See the parametric plot section. – WimC Sep 23 '14 at 05:32 — WimC, Sep 23 '14 at 05:32
write $z$ as $cosø +isinø$ and see what $e^z$ is – avz2611 Sep 23 '14 at 05:40 — avz2611, Sep 23 '14 at 05:40

score 2 · Accepted Answer · answered Sep 23 '14 at 10:26

Write $z=x+iy$ and the condition $|z|=1$ becomes $x^2+y^2=1$.

Now $e^z=e^{x+iy}=e^x e^{iy}=e^x(\cos y+i\sin y)=(e^x\cos y)+i(e^x\sin y)$. Put $x=\pm\sqrt{1-y^2}$ for each $-1\le y\le 1$ (two times, once with $+$, the other with $-$) and your graph becomes

$$f(y)=(e^{\pm\sqrt{1-y^2}}\cos y)+i(e^{\pm\sqrt{1-y^2}}\sin y)$$

The graph of $e^z$ for $|z|=1$

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