### Author Topic: Q3 TUT 5201  (Read 7589 times)

#### Victor Ivrii

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##### Q3 TUT 5201
« on: October 12, 2018, 06:21:02 PM »
Show that $w = \sin(z)$ maps the strip $\{x+yi\colon \pi/2 < x < \pi/2\}$ both one-to-one and onto the region obtained by deleting from the plane the two rays $(\infty, -1]$ and $[1, \infty)$.

Draw both domains.

Hint: Use equalities $\sin (-z)=-\sin(z)$, and $\overline{\sin(z)} = \sin(\bar{z})$.

#### Ende Jin

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##### Re: Q3 TUT 5201
« Reply #1 on: October 30, 2018, 09:33:50 PM »
Showing Bijective: We first show it is bijective between $\{x +yi : 0 < x < \frac{\pi}{2}, y > 0\}$ and $\{ai + b : a$>$0 , b$>$0 \}$ and because of symmetry ($\sin(\bar{x}) = \overline{\sin (x)}$ and $\sin(-x) = -\sin(x)$) we can get other quadrant for free (not including axis)\\

{When $a > 0, b > 0$}

Let $z = x+yi$ where $x, y \in \mathbb{R}$
\begin{align*}
\sin(z) & = \sin(x+yi) \\
& = i \cos x \frac{e^{y} - e^{-y}}{2} + \sin x \frac{e^{-y} + e^{y}}{2}
\end{align*}

Equivalently, we need to show, for all $a > 0, b > 0$
\begin{align*}
\cos x \frac{e^{y} - e^{-y}}{2}  &= a \\
\sin x \frac{e^{-y} + e^{y}}{2} &= b
\end{align*}
Has one and only one solution in $x \in (0, \frac{\pi}{2}), y > 0$.
We make a manipulation that,

from (Equation 1)$^2$ + (Equation 2)$^2$

and (Equation 1)$^2$ - (Equation 2)$^2$ we can get
\begin{align*}
\frac{e^{2y} + e^{-2y}}{4} - \frac{1}{2} \cos 2x &= a^2 + b^2 \\
\frac{e^{2y} + e^{-2y}}{4} \cos 2x - \frac{1}{2} &= a^2 - b^2
\end{align*}

Let $u = e^{2y} + e^{-2y}, v = \cos 2x$ (we can see that both $u,v$ are injective when $x \in (0, \frac{\pi}{2}), y > 0$),
we get equations
\begin{align}
\frac{u}{4} - \frac{v}{2} &= a^2 + b^2  \label{eq:3}\\
\frac{uv}{4} - \frac{1}{2} &= a^2 - b^2 \label{eq:4}
\end{align}
Thus, we only need to show $u,v$ has one and only one solution in the above equations where $u \in [2, \infty), v \in [-1,1]$
We can eliminate $u, v$ in \ref{eq:4} respectively by substituting from \ref{eq:3}. Thus we get,
\begin{align}
v^2 + 2(a^2+b^2)v - (2(a^2-b^2) + 1) &= 0 \\
u^2 - 4(a^2+b^2)u - (8(a^2-b^2) + 4) &= 0
\end{align}
define
\begin{align*}
f(v) &= v^2 + 2(a^2+b^2)v - (2(a^2-b^2) + 1) \\
g(u) &= u^2 - 4(a^2+b^2)u - (8(a^2-b^2) + 4)
\end{align*}

Since $f(-1) = -4a^2 < 0$ and $f(1) = 4b^2 > 0$ thus by intermediate theorem, there is one  solution for $v$ in (-1,1), and it is the only one in (-1,1) because it is a parabola.

Since $g(2) = -16a^2 < 0$ and $\lim_{x \rightarrow \infty} g(x) = \infty$, again by intermediate theorem, there is one solution for $u$ in (2, $\infty$], and it is the only one in  (2, $\infty$]  because it is a parabola.

{When $a = 0, 0 < b \le 1$}
Similarly as above, we get
\begin{align*}
\cos x \frac{e^{y} - e^{-y}}{2}  &= 0 \\
\sin x \frac{e^{-y} + e^{y}}{2} &= b
\end{align*}
Thus, $y = 0$, it is trivial to see there is a solution $\sin x = b$ since $b < 1$, it is the only one because $\sin$ is injective in $(-\frac{\pi}{2}, \frac{\pi}{2})$

{When $a > 0, b = 0$}
\begin{align*}
\cos x \frac{e^{y} - e^{-y}}{2}  &= a \\
\sin x \frac{e^{-y} + e^{y}}{2} &= 0
\end{align*}
Thus $x = 0$, since $\frac{e^{y} - e^{-y}}{2} = a \Leftrightarrow e^{2y} -2ae^{y} - 1$ where $\Delta = 4a^2 + 4 > 0$ thus we have a solution. It is the only solution because $\alpha \mapsto e^{\alpha} - e^{-\alpha}$ is an strictly increasing function (by derivative), which means injective.

We have shown there is one and only one solution in domain from different parts of the codomain. Thus it is bijective
« Last Edit: October 30, 2018, 09:37:47 PM by Ende Jin »

#### Victor Ivrii

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##### Re: Q3 TUT 5201
« Reply #2 on: October 31, 2018, 12:07:47 AM »
Where is drawing? And it would be simpler to use hyperbolic functions $\cosh(y)=\frac{e^y+e^{-y}}{2}$ and $\sinh(y)=\frac{e^y-e^{-y}}{2}$; then $\cos^2(y)-\sinh^2(y)=1$.

#### Ende Jin

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• Karma: 11
##### Re: Q3 TUT 5201
« Reply #3 on: October 31, 2018, 01:18:31 PM »
Can you elaborate that part using the identity of $\sinh, \cosh$? Where to use it?