### Author Topic: Q3 TUT 0201  (Read 1099 times)

#### Victor Ivrii ##### Q3 TUT 0201
« on: October 12, 2018, 06:15:42 PM »
$\renewcommand{\Im}{\operatorname{Im}}\renewcommand{\Re}{\operatorname{Re}}$
Show that $F(z) = e^z$ maps the strip
$$S = \{z=x + iy\colon -\infty < x < \infty, -{\pi}/{2} \le y \le {\pi}/{2}\}$$
onto the region $Q = \{w = s + it\colon s\ge 0, w \ne 0\}$ and that $F$ is one-to-one on$S$.

Furthermore, show that F maps the boundary of $S$ onto all the boundary of $Q$ except $w = 0$. Explain what happens to each of the horizontal lines $\{z\colon \Im z = \pi/2\}$ and $\{z\colon \Im z = -\pi/2\}$.

Draw both domains.

#### Junya Zhang

• Full Member
•   • Posts: 27
• Karma: 29 ##### Re: Q3 TUT 0201
« Reply #1 on: October 12, 2018, 06:26:46 PM »
This is question 23 from textbook section 1.5.

(1) Show that $F(z) = e^z$ maps the strip $S=\{x+iy:-\infty < x < \infty, -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\}$ onto the region $\Omega = \{w=s+it:s\geq 0, w \neq 0\}$.
Let $z=x+iy$ be a complex number in $S$. Then we know that $x\in(-\infty, \infty)$ and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$$F(z) = e^{x+iy} = e^xe^{iy} = e^x(\cos y + i\sin x) = e^x\cos y + ie^x\sin y$$
Let $F(z)=w=s + it$. Then $s = e^x\cos y$ and $t = e^x\sin y$.
Note that $\forall y \in [-\frac{\pi}{2}, \frac{\pi}{2}], \cos y \geq 0$ and $\forall x \in (-\infty, \infty), e^x > 0$, thus $s = e^x\cos y \geq 0$.
Moreover, $s = 0$ if and only if $\cos y = 0$, which happens when $y = \frac{\pi}{2}$ or $-\frac{\pi}{2}$. However, when $y = \frac{\pi}{2}$, $t = e^x\sin y = e^x \neq 0$ and when $y = -\frac{\pi}{2}$, $t = e^x\sin y = -e^x \neq 0$. This shows that $F(z) = w = s+it \neq 0$.
Thus, $F(z)\in \Omega$.
Since $z$ was an arbitrary element in $S$, this shows that $F$ maps the strip $S$ onto the region $\Omega$.

(2) Show that the $F$ is one-to-one on $S$.
To show that $F$ is one-to-one on $S$, we need to show that $\forall z_1, z_2 \in S, F(z_1)=F(z_2)\Rightarrow z_1=z_2$.
Let $z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$ be two arbitrary elements in $S$, and suppose $F(z_1)=F(z_2)$. Then by computation in part (1) we have $$F(z_1) = e^{x_1}\cos {y_1} + ie^{x_1}\sin {y_1}$$ $$F(z_2) = e^{x_2}\cos {y_2} + ie^{x_2}\sin {y_2}$$
Equating $F(z_1)$ and $F(z_2)$ we get $$e^{x_1}\cos {y_1} = e^{x_2}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_2}\sin {y_2}$$
Square both sides of the two equations we get $$e^{2x_1}\cos^2 {y_1} = e^{2x_2}\cos^2 {y_2}$$ $$e^{2x_1}\sin^2 {y_1} = e^{2x_2}\sin^2 {y_2}$$
Add the two equations together we get $$e^{2x_1}(\cos^2 {y_1} + \sin^2 {y_1}) = e^{2x_2}(\cos^2 {y_2} + \sin^2 {y_2} )$$
Simplify this we get $e^{2x_1} = e^{2x_2}$. Since $e^{2x_2} \neq 0$, we can divide both sides of the equation by $e^{2x_2}$ and get $e^{2x_1 - 2x_2} = 1$, which implies $2x_1 - 2 x_2 = 0$. Therefore we can conclude $x_1 = x_2$.

Back to the equations $$e^{x_1}\cos {y_1} = e^{x_2}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_2}\sin {y_2}$$
Now substitute in $x_1 = x_2$ we have $$e^{x_1}\cos {y_1} = e^{x_1}\cos {y_2}$$ $$e^{x_1}\sin {y_1} = e^{x_1}\sin {y_2}$$
This implies $$\cos {y_1} = \cos {y_2}$$ $$\sin {y_1} = \sin {y_2}$$
Multiply the first equation by $\sin {y_2}$ and the second equation by $\cos {y_2}$ we get $$\sin {y_2}\cos {y_1} = \sin {y_2}\cos {y_2}$$ $$\sin {y_1}\cos {y_2} = \sin {y_2}\cos {y_2}$$
Subtract the second equation from the first we get $$\sin {y_2}\cos {y_1} - \sin {y_1}\cos {y_2} = \sin {y_2}\cos {y_2} - \sin {y_2}\cos {y_2}$$
Simplify this we get $$\sin ({y_2}- {y_1})= 0$$
This implies ${y_2}- {y_1} = k\pi$ for $k\in\mathbb{Z}$.
Since $y_1,y_2\in [-\frac{\pi}{2}, \frac{\pi}{2}]$, then $y_2 - y_1 \in [-\pi,\pi]$.
Together, possible values for $y_2 - y_1$ are $-\pi, 0, \pi$.
If $y_2 - y_1 =-\pi$, then $y_2 = -\frac{\pi}{2}$ and $y_1 = \frac{\pi}{2}$.
If $y_2 - y_1 =\pi$, then $y_1 = -\frac{\pi}{2}$ and $y_2 = \frac{\pi}{2}$.
From $\sin {y_1} =\sin {y_2}$ we can eliminate these two cases since $\sin(-\pi/2) = -1$ and $\sin(\pi/2) = 1$
Therefore, it must be the case that $y_1 - y_2 = 0$. That is $y_1 = y_2$.

Since $x_1 = x_2$ and $y_1 = y_2$, we can conclude that $z_1 = z_2$. This shows that $F$ is one-to-one on $S$.

(3) Show that $F$ maps the boundary of $S$ onto all the boundary of $\Omega$ except $w=0$.
The boundary of $S$ is the set $\{x+iy : -\infty < x < \infty, y = \frac{\pi}{2} \text{ or } y = -\frac{\pi}{2}\} = \{z: \text{Im }z = \frac{\pi}{2}\} \cup \{z: \text{Im }z = \frac{-\pi}{2}\}$ and the boundary of $\Omega$ is the set $\{s+it:s = 0\}$
We will show in part (4) that $F$ maps the set $\{z: \text{Im }z = \frac{\pi}{2}\}$ to $\{s+it:s = 0, t > 0\}$ and maps the set $\{z: \text{Im }z = \frac{-\pi}{2}\}$ to $\{s+it:s = 0, t < 0\}$, which we can then conclude that $F$ maps the boundary of $S$ onto the boundary of $\Omega$ except for the origin.

(4) Explain what happens to each of the horizontal lines $\{z: \text{Im }z = \frac{\pi}{2}\}$ and $\{z: \text{Im }z = -\frac{\pi}{2}\}$.
Let $z=x+i\pi/2$ be a complex number on the line $\{z: \text{Im }z = \frac{\pi}{2}\}$. The function $F(z)=e^z$ maps $z$ to $e^x\cos \pi/2 + ie^x\sin \pi/2 = ie^x$. Since the function $f(x) = e^x$ is a one-to-one and onto function from $(-\infty, \infty)$ to $(0,\infty)$, this shows that $F$ maps the line $\{z: \text{Im }z = \frac{\pi}{2}\}$ onto the positive imaginary axis (not including the origin).
Similarly, let $z=x+i(-\pi/2)$ be a complex number on the line $\{z: \text{Im }z = \frac{-\pi}{2}\}$. The function $F(z)=e^z$ maps $z$ to $e^x\cos (-\pi/2) + ie^x\sin (-\pi/2) = -ie^x$. Since $f(x) = -e^x$ is a one-to-one and onto function from $(-\infty, \infty)$ to $(-\infty, 0)$, this shows that $F$ maps the line $\{z: \text{Im }z = -\frac{\pi}{2}\}$ onto the negative imaginary axis (not including the origin).

(5) Draw both domains.
See image from textbook page 55.
« Last Edit: October 12, 2018, 07:36:51 PM by Junya Zhang »