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Consider the strip in the complex plane defined by $\{z\in\mathbb{C}:\Im(z)\in (0,\pi)\}$. Applying $\operatorname{exp}$ to this strip maps it onto the half plane $\{z\in\mathbb{C}:\Im(z)>0\}$?

What is a formal explanation why, other than just plotting a few points? Does the width of the strip matter, or does $\operatorname{exp}$ map similar strips, like $\{z\in\mathbb{C}:\Im(z)\in (0,1)\}$ to half planes also?

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1 Answer 1

Since

$$\exp(x+iy) = \exp(x)\exp(iy) = \exp(x)(\cos y + i\sin y)$$

The strip $\{x+iy : l\leq y\leq k\}$ is mapped by exp to a cone (or a sector) which extends radially to infinity, and whose angular extent is determined by the limits on $y$.

For the special case $0\leq y\leq \pi$ this represents a half-plane, since cos maps the interval $[0,\pi]$ to $[-1,1]$, but sin maps $[0,\pi]$ only onto $[0,1]$ (i.e. the imaginary part must be positive).

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