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And when you put that into the nth root form... It becomes $2^{1/18}\cos\theta + 2^{1/18}\sin\theta$?

$n$th root form given is: $\sqrt[n]r\cdot\cos(\theta+2\pi k)n$

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4 Answers 4

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You have $\theta = \frac{3\pi}{4}$ and $r=\sqrt{2}$, and using the fact that $e^{i2k\pi}=1$, then

$$ z^3 = (-1+i) = \sqrt{2}\,e^{i\frac{3\pi}{4}}= \sqrt{2}\,e^{i\frac{3\pi}{4}}e^{i2k\pi}=\sqrt{2}e^{i\frac{3\pi}{4}+i2k\pi} $$

$$ \implies z = 2^{1/6}\,e^{\frac{i\pi}{4}+\frac{i2k\pi}{3}} \,. $$

Now, taking $k=0,1,2$ gives the three roots. Note that, if you take $k>2$ you will get the same roots.

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Think of $-1 + i = r e^{i \theta}$ geometrically as the point $(-1,1)$ so it has length $r = \sqrt{2}$.

So the cube root of it has length $\sqrt[3]{r} = \sqrt[3]{\sqrt{2}} = \sqrt[6]{2}$.

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Okay, but when I try to find the nth roots using the formula given, does it not turn it into 2^1/18? –  David Thompson Nov 1 '12 at 23:22
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The easiest if you know the exponential function: $e^{x+iy}=e^x(\cos y+i\sin y)$.

Then, if you drawn in coordinate system, you will find the angle $\vartheta$ of $-1+i$. Then, $$-1+i= \sqrt2\cdot e^{\theta i} $$ this makes easier the power to $1/3$. Don't forget that you will have 3 solutions, as $e^{4\pi i}=e^{2\pi i}=1$ but their thirds are different.

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$(1+i)^\frac{1}{3} = (|1+i| e^{i \arg (1+i)})^\frac{1}{3} = (\sqrt{2} e^{i \frac{\pi}{4}})^\frac{1}{3}$

The $3$rd roots are given by $(\sqrt{2} e^{i (\frac{\pi}{4}+2 \pi k)})^\frac{1}{3} = 2^{\frac{1}{6} }e^{i (\frac{\pi}{12}+ \frac{2\pi k}{3})} = 2^{\frac{1}{6} }(\cos (\frac{\pi(1+8k)}{12}) + i \sin \frac{\pi(1+8k)}{12})$, for $k=0,1,2$.

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