# 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

-
Use $i = e^{i \frac{\pi}{2}}$ to expand the powers of $i$, and then use Euler's formula. – copper.hat Feb 5 '13 at 22:32
@copper.hat, how can this be? can you please explain me with small proof? btw, i forgot $e$ in my question, now updated – doniyor Feb 5 '13 at 22:33
There is the famous equation $e^{\pi i}+1=0$ – Hagen von Eitzen Feb 5 '13 at 22:34
Are you familiar with $e^{i \theta} = \cos \theta + i \sin \theta$? – copper.hat Feb 5 '13 at 22:34
@copper.hat, oh yeah. now i see what you mean – doniyor Feb 5 '13 at 22:35

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):

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

-
Thanks, but i forgot $e$ then updated my question, can you please update your answer also? – doniyor Feb 5 '13 at 22:36

Use Euler's formula:

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

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

-
Thanks, yeah now i got – doniyor Feb 6 '13 at 22:16

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

-
Thanks anorton, great – doniyor Feb 6 '13 at 22:15

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

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.

-
Georg, very nice. thanks in tons dude. – doniyor Feb 6 '13 at 22:15