# Exponential function and bijection

I am required to prove the following:

For any real number $k$, prove that the exponential function $e^z$ is a bijection ($z$ is a complex number) from the strip $a < im z \leq k+2pi$ to the complex plane minus the point $0$, $\Bbb C - \{0\}$.

Any hints please? Thanks!

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Michael, yup, aa was intended to be k, fell asleep on the keyboard:p – PongYoPongYo Aug 26 '12 at 14:20
It can't be any $k\in \mathbb{R}$, because what if $k=3$ and $im(z)=2$? – MBP Aug 26 '12 at 17:46

Hint To solve $e^{x+iy}=\omega$, write $\omega$ in trigonometric form and solve.
Given a $\omega$, you can solve $\omega=e^x(\cos y+i\sin y)$ for $x$ and $y$. You can easely find $x$ by taking the absolute value, and then prove that there is only one $y$ in your domain. – N. S. Aug 26 '12 at 14:44
A hint: Don't solve equations, but investigate what the exponential function $$z=x+iy \ \mapsto\ e^z=e^x\cdot e^{iy}$$ does to horizontal lines $$g_v:\quad y:= v\ (={\rm const.})\ , \quad -\infty<x<\infty\ ,$$ and to vertical lines $$h_u:\quad x:= u\ (={\rm const.}), \quad -\infty<y<\infty\ .$$