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!

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 $$r=\sqrt[\,2]{x^2+y^2}\qquad \mbox{and } \qquad \tan(\theta)=\frac{y}{x},\quad -\frac{\pi}{2}<\theta<\frac{\pi}{2}$$

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.

Recall Euler's Formula:

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

By applying this formula to the given problems we obtain,

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