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$(2i)^{1/2}$

$(1-\sqrt{3}i)^{1/2}$

$(-1)^{1/3}$

$(-16)^{1/4}$

How can I find the roots of the complex numbers above?

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May look into math.stackexchange.com/questions/201769/… –  lab bhattacharjee Oct 25 '12 at 8:30

2 Answers 2

up vote 2 down vote accepted

Use de Moivre's theorem: $$z^{1/n} = r^{1/n} \exp \left( i\frac{\theta + 2\pi k}{n} \right)$$ for $k = 0, 1, 2, \ldots, n - 1$. The formula is not difficult to derive as we only need to write $z$ in the polar form and take the $n$th root of both sides.

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How do I find r, n, and theta here? –  user1038665 Oct 25 '12 at 2:00
    
Here $r$ is the radius, $\theta$ is the angle and $n$ depends on the problem. In general we seek the $n$-th root of a complex number. For example, for ${(2i)^{1/2}}$ we see that $n = 2$ because we want the square root of $2i$. We then use Argand diagram to find $r$ and $\theta$, that is, we represent a complex number as a vector. In this case $r = 2$ and $\theta = \pi /2$ because $2i$ is essentially a horizontal line of length 2. –  glebovg Oct 25 '12 at 2:10
    
In other cases we use the Pythagorean theorem to find $r$, and $\theta$ comes from our knowledge of common angles (usually). Try to find $r$ and $\theta$ for the rest of the complex numbers. –  glebovg Oct 25 '12 at 2:17
    
If you still do not understand use YouTube, there must be a good explanation. –  glebovg Oct 25 '12 at 2:18

Use DeMoivre's formula.

Don't forget that cos and sin functions are $2\pi$ periodic

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