Author Topic: TT1 Problem 5 (noon)  (Read 1153 times)

Victor Ivrii

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TT1 Problem 5 (noon)
« on: October 19, 2018, 04:09:49 AM »
Find any region that is mapped bijectively to $\{w\colon \Re w\ge -|\Im w |, \ 0< |w|\le 2\}$ by the map $w=e^{iz}$. Draw both of them.
« Last Edit: October 19, 2018, 04:19:11 AM by Victor Ivrii »


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Re: TT1 Problem 5 (noon)
« Reply #1 on: October 20, 2018, 05:27:00 PM »
The domain region is the region $\Im z \geq -\ln 2, \Re z \in [-\frac{3\pi}{4}, \frac{3\pi}{4}]$, or equivalently $y \leq \ln 2, x \leq \frac{3\pi}{4}$.

The range region is contained within the circle $|w| \leq 2$. The $\Re w \geq -|\Im w|$ can be viewed as $x \geq -|y|$. That means, it is the region to the right of line $x \geq -y$ for the top half, and $x \geq +y$ for the bottom half.

The range region must have an absolute value of Arg less or equal to $\frac{3\pi}{4}$, and a magnitude less or equal to $2$.
It can be described as $\{w | \frac{3\pi}{4} \leq \operatorname{Arg}w \leq \frac{3\pi}{4} \wedge 0 < |z| \leq 2\}$

The function $e^{iz} = e^{i(x + iy)} = e^{-y + ix}$. Hence, the x-coordinate determines the Argument, and the y-coordinate determines the magnitude (increases as y moves downwards).
Hence, the function $e^{iz} = (e^{-y})(\cos x + i \sin x)$.

Prove that this function is injective
For $\Im  z \geq -\ln 2$, it follows $-y \leq \ln 2$, and so the magnitude $e^{-y} \leq 2$.
For $-\frac{3\pi}{4} \leq \Re x \leq \frac{3\pi}{4}$, the argument of $w$ will be in $[-\frac{3\pi}{4}, \frac{3\pi}{4}]$
Two nonzero complex numbers are the same when both the magnitude and principal Argument are the same. The function $e^{iz}$ transforms the real and imaginary parts of $z$ into the argument and the reciprocal magnitude of the output. If the outputs are the same (Arg, magnitude), the inputs are the same ($x$, $-y$). Hence, this function is injective.

Prove that this function is surjective
Given any complex number with $0 < |z| \leq 2$, and $ -\frac{3\pi}{4} \leq \operatorname{Arg}w \leq \frac{3\pi}{4}$. The input $x + iy$ would have the imaginary component (encodes the reciprocal of the magnitude) to be $e^-y \leq \ln 2$, so $y \leq \ln 2$, and finally $-y \geq \ln 2$, and the real component (encodes the Argument) to be in $ -\frac{3\pi}{4} \leq x \leq \frac{3\pi}{4}$
« Last Edit: October 20, 2018, 09:43:41 PM by oighea »