Find the proof of the argument
To start , let's expand the exponential function
$z=x+iy$, putting $x=0$, we get $z=iy$
$e^{iy}=1+\frac{iy}{1!}+\frac{{iy}^{2}}{2!}+ \frac{{iy}^{3}}{3!}+ \frac{{iy}^{4}}{4!}+...$
$=(1-\frac{y^2}{2!}+\frac{y^4}{4!}-... )
+i(y-\frac{y^3}{3!}+\frac{y^5}{5!}-... )$
$=cosy+isiny$
$e^{z}=e^{x}e^{iy}=e^{x}(cos (y)+isin (y))$
$x+iy=\gamma(cos\theta -isin\theta)=\gamma e^{i\theta}$
$z=x+iy[\textbf{Cartesian form}]$
$=\gamma(sin\theta +isin\theta)[\textbf{Polar form}]$
$=\gamma e^{i\theta}[\textbf{Exponential form}]$
$-1=-1+i(0)$
$\implies \gamma=\sqrt(-1)^{2}=1$
$\implies \theta =\pi$
$\implies e^{i\pi}=-1$
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