MAT334-2018F > Quiz-3

Q3 TUT 5201

(1/1)

Victor Ivrii:
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:
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

Victor Ivrii:
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:
Can you elaborate that part using the identity of $\sinh, \cosh$? Where to use it?

Victor Ivrii:
You just wrote expressions for sinh, cosh without calling them this way