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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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