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Euler's identity
Solution
To start, let’s expand the exponential function
\[e^{z}=1+\frac{z}{1 !}+\frac{z^{2}}{2 !}+\ldots+\frac{z^{n}}{n !}+\cdots\]
\(z=x+iy\), putting \(x=0\) we get \(z=iy\)
\begin{aligned}
e^{i y} &=1+\frac{i y}{1 !}+\frac{(i y)^{2}}{2 !}+\frac{(i y)^{3}}{3 !}+\frac{(i y)^{4}}{4 !}+\cdots \\
&=\left(1-\frac{y^{2}}{2 !}+\frac{y^{4}}{4 !}-\cdots\right)+i\left(y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots\right) \\
&=\cos y+i \sin y
\end{aligned}
\[e^{z}=e^{x} e^{i y}=e^{x}(\cos y+i \sin y)\]
\[\begin{aligned}
x+i y &=\gamma(\cos \theta+i \sin \theta)=\gamma e^{i \theta} \\
z &=x+i y \quad[\text { cartesian form ] }\\
&=\gamma(\sin \theta+i \sin \theta) \quad[\text { polar form ]} \\
&\left.=\gamma e^{i \theta} \quad \text { [ exponential form }\right]
\end{aligned}\]
\[\begin{aligned}
&-1=-1+i(0) \\
&\Rightarrow \gamma= \sqrt{(-1)^{2}}=1 \\
&\Rightarrow \theta=\pi \\
&\Rightarrow e^{i\pi}=-1
\end{aligned}
\]
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