rewrite $2ie^{i\pi}+i^3$ i am asked to rewrite $2ie^{i\pi}+i^3$ into $x+iy$ form. i just tried all what i know so far, but couldnot come to solution. i said: $2ie^{i\pi}+i^3=2ie^{i\pi}-i$ but further i am stuck really. i am really eager to learn how things like this work. i appreciate any help and attempt to help. 
the more difficult problems i am facing, the more i am loving maths. this problem was the first problem in my exam today. it took me 20 minutes. no sign of success..  
EDIT: sorry, i forgot $e$. now it is there
 A: Using the main branch for the logarithmic function
$$i^{\pi i}=e^{\pi i\operatorname{Log}(i)}=e^{\pi i\left(\log|i|+\frac{\pi i}{2}\right)}=e^{\pi i\frac{\pi i}{2}}=e^{-\frac{\pi^2}{2}}\Longrightarrow$$
$$2i^{\pi i}+i^3=2e^{-\frac{\pi^2}{2}}-i$$
Added: after a modification of the question by the OP (much simpler now):
$$2ie^{\pi i}+i^3=-2i-i=-3i$$
A: Use Euler's formula:
$$e^{ix} = cos(x) + i sin(x)$$.
Plug in $x = \pi$ and you could then move on.
A: $$2ie^{i\pi}+i^3$$
Recall that:
$$e^{i\theta} = \cos\theta + i\sin\theta$$
Thus, $e^{i\pi} = -1$.  Substituting:
$$-2i+i^3$$
Continuing using basic properties of $i$:
$$-2i-i$$
$$\boxed{-3i}$$
A: $$ 2ie^{i\pi} + i^3 $$
$$ 2ie^{i\pi} - i $$
Recall Euler's Identity:
$$ e^{i\pi} + 1 = 0$$
and so:
$$ e^{i\pi} = -1 $$
$$ 2i \cdot -1 - i$$
$$ -2i - i $$
$$ \color{red}{-3i} $$

Alternatively, use Euler's Formula, that:
$$ e^{ix} = \cos x + i \sin x $$
Using $x=\pi$:
$$ e^{i\pi} = \cos \pi + i \sin \pi $$
$$ e^{i\pi} = -1 + 0i $$
$$ e^{i\pi} = -1 $$
And proceed as above.
