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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?

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answer =$-1+i$ arguement = $\frac{3\pi}{4}$ –  ProMatheus Nov 12 '13 at 19:38

2 Answers 2

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 $0$).

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Hint:

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

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