Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Use De Moivre's Theorem to determine $(-1 +i)^{184}$ in the form $x + iy$

I first rewrite the equation in polar form.

To do this I first determine $z$
$z = -1 + i$ I then solve
$|z| = \sqrt{-1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$

I now determine theta by getting the arctan

$\theta = tan^{-1}\frac{-1}{1} = -1 = -\frac{\pi}{4} $

Use of a calculator is not permitted in the exam, but I understand that $tan^{-1} 1 = \frac{\pi}{4}$ is just one of those things that you need to remember.

I now write my equation in polar form
$[\sqrt{2} (cos-\frac{\pi}{4} + isin-\frac{\pi}{4})]^{184}\\ =(\sqrt{2})^{184} (cos-\frac{\pi}{4} + isin-\frac{\pi}{4})^{184}\\ = 16 (cos(-46\pi) + isin(-46\pi))$

But here I am stuck. How do I proceed from this point, assuming what I have done so far is correct?

share|cite|improve this question
answer =$-1+i$ arguement = $\frac{3\pi}{4}$ – Suraj M S Nov 12 '13 at 19:38

Be careful with arctangent; remember that $\tan^{-1}$ sometimes gives you an answer that is in the wrong quadrant. Sketch the point $-1 + i$ on the plane and decide if $-\pi/4$ is really the angle you want.

After that, proceed as you have, and finally, you just need to simplify expressions like $\cos(k\pi)$ and $\sin(k\pi)$ for an integer $k$. You should know how to do this (hint: they're equal to $1,-1$ or $0$).

share|cite|improve this answer


$$ \cos(2n\pi)=1,\quad n=0,\pm 1,\pm 2\dots,\quad \sin(2n\pi)=0,\quad n=0,\pm 1,\pm 2\dots. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.