### Author Topic: TT1 Problem 1 (morning)  (Read 2350 times)

#### Victor Ivrii

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##### TT1 Problem 1 (morning)
« on: October 19, 2018, 03:51:50 AM »
Show that
\begin{equation*}
|\cos (z)|^2 = \cos^2 (x)+ \sinh^2 (y)
\end{equation*}
for all complex numbers $z = x+yi$.

#### Meng Wu

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• MAT3342018F
##### Re: TT1 Problem 1 (morning)
« Reply #1 on: October 19, 2018, 07:26:54 AM »
\begin{align}\cos(z)=\cos(x+iy)&=\cos(x)\cos(iy)-\sin(x)\sin(iy) \\&= \cos(x)\cosh(y)-i\sin(x)\sinh(y)\end{align}
\require{cancel}\begin{align}|\cos(z)|^2=|\cos(x+iy)|^2&=\Bigg(\sqrt{\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y)}\Bigg)^2 \\ &=\cos^2(x)\cosh^2(y)+\sin^2(x)\sinh^2(y) \\&= \cos^2(x)(1+\sinh^2(x))+(1-\cos^2(x))\sinh^2(y)\\&=\cos^2(x)+\cancel{\cos^2(x)\sinh^2(x)}+\sinh^2(y)-\cancel{\cos^2(x)\sinh^2(y)}\\&=\cos^2(x)+\sinh^2(y)\end{align}
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#### Meng Wu

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• MAT3342018F
##### Re: TT1 Problem 1 (morning)
« Reply #2 on: October 19, 2018, 07:50:22 AM »
$\textbf{Alternative Method}:$
$$\\$$
Since,
$$z=x+iy, \text{ where } x,y \in \mathbb{R}.$$
$$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$$
Then,
\begin{align}\cos(z)=\cos(x+iy)&=\frac{e^{i(x+iy)}+e^{-i(x+iy)}}{2}\\&= \frac{e^{-y+ix}+e^{y-ix}}{2}\\&=\frac{e^{-y}[\cos(x)+i\sin(x)]+e^{y}[\cos(-x)+i\sin(-x)]}{2}\\&=\frac{e^{-y}\cos(x)+ie^{-y}\sin(x)+e^{y}\cos(x)-e^{y}\sin(x)}{2}, \\\text{since } \cos(-x)=\cos(x) \text{ and }\sin(-x)=-\sin(x).\\&=\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2}\end{align}
Hence,
\begin{align}|\sin(z)|=|\sin(x+iy)|&=\Bigg|\frac{\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)}{2} \Bigg|\\&=\frac{\Bigg|\cos(x)(e^{-y}+e^y)+i\sin(x)(e^{-y}-e^y)\Bigg|}{\Big|2\Big|}\\&=\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\end{align}
Thus,
\begin{align}|\sin(z)|^2=|\sin(x+iy)|^2&=\Bigg(\frac{\sqrt{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}}{2}\Bigg)^2\\&=\frac{[\cos(x)(e^{-y}+e^y)]^2+[\sin(x)(e^{-y}-e^y)]^2}{4}\\&=\frac{[\cos^2(x)(e^{-2y}+e^{2y}+2)]+[\sin^2(x)(e^{-2y}+e^2y-2)]}{4}\\&=\frac{e^{-2y}(\cos^2(x)+\sin^2(x))+e^{2y}(\cos^2(x)+\sin^2(x))+2(\cos^2(x)-\sin^2(x))}{4}\\&=\frac{e^{-2y}+e^{2y}+2(2\cos^2(x)-1)}{4}\\&=\frac{e^{-2y}+e^{2y}-2+4\cos^2(x)}{4}\\&=\frac{e^{-2y}+e^{2y}-2}{4}+\frac{4\cos^2(x)}{4}\\&=\sinh^2(y)+\cos^2(x)\end{align}
Note:
$$\sinh(y)=\frac{e^{y}-e^{-y}}{2}, \text{ where } y\in \mathbb{R}.$$
\begin{align}\sinh^2(y)&=\Bigg(\frac{e^{y}-e^{-y}}{2}\Bigg)^2\\&=\frac{e^{2y}+e^{-2y}-2}{4}\end{align}
Therefore, we have proven $|\cos(z)|^2=\cos^2(x)+\sinh^2(y)$ for all $z=x+iy$.