Exponential to polar form

Question

I have exponential form $$ je^{-j\pi/2} $$, where $j = \sqrt{-1}$

I want to convert this to polar form $$j(\cos\pi/2 + j \sin \pi/2)$$

is it correct?

Answer 1

Don't forget that negative.

$$je^{-j\frac\pi2}$$

$$=j[cos(-\frac\pi2) + jsin(-\frac\pi2)]$$

$$=j[cos(\frac\pi2) - jsin(\frac\pi2)]$$

Answer 2

$$je^{-j\pi/2} =j(\cos(-\pi/2) +j\sin(-\pi/2))=j(\cos \pi/2 -j\sin \pi/2)=j(0-j)=-j^2=1$$

Alternatively, since $j=e^{j\pi/2}$,

$$je^{-j\pi/2}=e^{j\pi/2}e^{-j\pi/2}=e^{-j\pi/2+j\pi/2}=e^0=1$$