Expressing complex numbers in form $a+bi$ I know that we should express complex numbers generally in the standard form 

$$a+bi:a,b\in\mathbb{R}$$

Like $4+5i-2=2+5i$.
But how do I express complex numbers like $e^{-i\pi/2}$ or $i+e^{2\pi i}$?
Thank you!
 A: Hint Use the Euler's Formula
$$
 e^{i\theta}=\cos(\theta)+i\sin(\theta).
$$
For exemple, 
$$
e^{-i\pi/2} =
\cos(-\pi/2)+i\sin(-\pi/2)
=
0+i\cdot (-1)
=
-i 
$$
For reverse process  $x+iy= r\cdot e^{i\theta}$ use the formulas
$$
r=\sqrt[\,2]{x^2+y^2}\qquad \mbox{and } \qquad \tan(\theta)=\frac{y}{x},\quad -\frac{\pi}{2}<\theta<\frac{\pi}{2}
$$
A: Recall Euler's Formula:

$e^{ix}=\cos x + i \sin x$.

By applying this formula to the given problems we obtain,


*

*$e^{-i \pi / 2} = \cos (-\frac{\pi}{2}) + i \sin( -\frac{\pi}{2}) = 0 + i(-1)=-i$.

*$i+e^{2 \pi i} = i + \cos(2 \pi) + i \sin(2 \pi) = i + 1 + i(0) = 1 + i.$

A: Well, every complex number has a modulus and an angle which corresponds to $r, \theta$ in $re^{i\theta}$. In short, for a complex number $z=a+ib$, we have $r=\sqrt{a^2+b^2}$,  $\theta=\tan^{-1}\frac{b}{a}$. To change it back, all you have to do is to use $a=r\cos\theta, b=r\sin\theta$.
In your question, you can choose to change either the polar form into the Cartesian or vice versa and you will get your answer.
