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Exam time tomorrow and I am not entirely sure if I am doing this right.

I first write -8 as a complex number
$z^3 = -8 = -8 \times 0i$

Calculate the modulus of z
$|z| = \sqrt{-8^2} = 8$

Get the arg of z
$tan^{-1} = \frac{0}{-8} = 0 = \pi$

Write the number in polar form $\theta = \pi + 2k\pi$
$z^3 = 8(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))$

Use De Moivre's theorem
$z = 8^\frac{1}{3}(cos(\pi + 2k\pi) + isin(\pi + 2k\pi))^\frac{1}{3}\\ = 2(cos(\frac{\pi}{3} + \frac{2k\pi}{3}) + isin(\frac{\pi}{3} + \frac{2k\pi}{3}))$

and then I fill out the equation with different values of k
$ k = 0, z= 2(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))\\ k = 1, z= 2(cos(\pi) + isin(\pi))\\ k = 2, z= 2(cos(\frac{5\pi}{3}) + isin(\frac{5\pi}{3})) $

Is this correct?

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Looks correct as far as it goes -- but you should probably note that each of the trigonometric values in your results have expressions in terms of rationals and square roots, and use those representations instead. (This is not always the case, though). –  Henning Makholm Nov 14 '13 at 18:02
    
My biggest concern is that I have not done trig in a decade, I actually can't remember any of these values, I keep having to reference them –  Leon Nov 14 '13 at 18:08
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Leon - if you look at the unit circle posted below, you'll see that the main angles you'll want to know (in radians) are the sines and cosines of $\theta = \pi/6,\;\theta = \pi/4,\; \theta = \pi/3$, and of course, $\theta = \pi/2,\;\pi,\;3\pi/2, \text{ and }\; \theta = 0 = 2k\pi$. Other multiples of those angles are simply a matter of locating the quadrant and determining the correct sign. –  amWhy Nov 14 '13 at 18:34
    
That's a great help –  Leon Nov 14 '13 at 19:16

1 Answer 1

up vote 4 down vote accepted

Looks good so far.

What you'll also likely want to do is evaluate $\cos \theta, \sin\theta$ for each $\theta.$ The trigonometric values needed here are for angles that are fairly basic, so you'll want to re-familiarize yourself with the standard angles, expressed in radians, and the corresponding trig values with those angles as arguments. See this link, and the image below, both from Wikipedia.

Taking your final work, but evaluating the $\sin$ and $\cos$ of the angles gives us an explicit representation of the cube roots of $-8$: $$ k = 0, z= 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 2\left(\frac 12 + i\frac {\sqrt 3}{2}\right) = 1 + i\sqrt 3\\ k = 1, z= 2(\cos(\pi) + i\sin(\pi)) = 2\left(-1 + i\cdot 0\right) = -2\\ k = 2, z= 2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}))= 2\left(\frac 12 - i\frac{\sqrt 3}{2}\right) = 1 - i\sqrt 3 $$

Trigonometric values of standard angles, using the unit-circle definition:

enter image description here

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That is such a nice picture +1 –  Amzoti Nov 15 '13 at 2:56

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