Question

$z=e^{i+i^2}$, find
$\left|z\right|=$?
argument =?

Answer 1

$e^{i+i^2}=e^{i-1}$

$=\frac{e^i}{e^1}$

$z=\frac{1}{e}[\cos 1+i\sin 1]$

[By demonrier theorem $e^{i\theta }=\cos \theta +i\sin \theta$]

$|z|=\frac{1}{e}\sqrt{{\cos }^{2}1+{\sin }^{2}1}$

$|z|=\frac{1}{e}$

$Tan\alpha =\frac{\sin 1}{\cos 1}=Tan1$

$\alpha =1$ radian

$={57.29}^o$

[ $\because 2\pi$ radians is ${180}^{\circ }$]

$1$ radian is $\frac{{180}^{\circ }}{2\pi }$