rewrite $2ie^{i\pi}+i^3$

Question

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

Answer 1

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):

$$2ie^{\pi i}+i^3=-2i-i=-3i$$

Answer 2

Use Euler's formula:

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

Plug in $x = \pi$ and you could then move on.

Answer 3

$$ 2ie^{i\pi} + i^3 $$ $$ 2ie^{i\pi} - i $$

Recall Euler's Identity:

$$ e^{i\pi} + 1 = 0$$ and so: $$ e^{i\pi} = -1 $$ $$ 2i \cdot -1 - i$$ $$ -2i - i $$ $$ \color{red}{-3i} $$

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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.

Answer 4

$$2ie^{i\pi}+i^3$$ Recall that: $$e^{i\theta} = \cos\theta + i\sin\theta$$ Thus, $e^{i\pi} = -1$. Substituting: $$-2i+i^3$$ Continuing using basic properties of $i$: $$-2i-i$$ $$\boxed{-3i}$$