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Find the image of $\{z: |\Im(z)| < \frac{\pi}{2}\}$ under the exponentional function.

So, i've set $e^{z} = e^x(\cos y + i\sin y)$ where $\Im(z) = e^x \sin y$

So I have to find the "image" such that $e^x \sin y < \frac{\pi}{2}$

Am I on the right track? I'm a bit confused as to where to go from here...

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You should not find the points where $e^x\sin y < \frac{\pi}{2}$. For the strip $S=\{ z = x+iy : \lvert y\rvert < \frac{\pi}{2}\}$, which region is the image of $S$ under the exponential function. So describe $\{ e^x(\cos y + i\sin y) : \lvert y\rvert < \frac{\pi}{2}\}$. – Daniel Fischer Mar 14 '14 at 19:44
I think you just rewrote it differently right? But we're both saying that the image would be $ - \frac{\pi}{2} < e^x siny < \frac{\pi}{2}$, since it's only asking for the imaginary part. I am not overly enlightened with abstrac/high level math, playing catch up here. – Blondie Mar 14 '14 at 19:51
@Blondie No, you are defining the preimage and not the image. See my answer for a concrete definition of what you're supposed to find. – AlexR Mar 14 '14 at 19:53
up vote 0 down vote accepted

The image of a set under function is defined by $$f(S) := \{f(x) | x\in S\}$$ Thus you want to find $$A:=\{e^{\Re z} (\cos \Im z + i \sin \Im z) : |\Im z| < \frac\pi2\}$$ Since $\Re z$ is arbitrary, you'll have "rays" from the origin, but not including the origin. The valid angles are between $-\frac\pi2$ and $\frac\pi2$ corresponding to the right half-pane. Note the strict inequality excludes the imaginary axis. Thus $$A = \{z\in\mathbb C : \Re z > 0\}$$

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Answer. The image of $\{z:\lvert{\mathcal Im}\,z\rvert<\pi/2\}$ is $\{z: {\mathcal Re}\,z>0\}$.


a. If $\lvert{\mathcal Im}\,z\rvert<\pi/2$, then $$ {\mathcal Re}\exp z={\mathcal Re}\exp (x+iy)=\exp (x)\cos y>0. $$

b. If ${\mathcal Re}\,z>0$, then $z=\varrho\mathrm{e}^{i\vartheta}$, with $\varrho>0$ and $\vartheta\in(-\pi/2,\pi/2)$, and hence $$ z=\exp(\log\varrho+i\vartheta). $$

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