Root of a complex number

1. Jun 19, 2007

kasse

z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

2. Jun 19, 2007

malawi_glenn

Have you tried De Moivre's formula ?

3. Jun 19, 2007

CompuChip

I believe the complex root is defined by (that is: it's usually continued from the real function by)
$$\sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2}$$

4. Jun 19, 2007

malawi_glenn

yes, and from De Moivre's formula we get the general:

$$z^{1/n} = r^{1/n}*exp(i \phi / n)$$

5. Jun 19, 2007

kasse

So e^((i Pi)/12) is a solution. How about the other three?

6. Jun 19, 2007

malawi_glenn

Remember that the same complex number can be obtained by rotating 2*pi and so on.

i.e arg((1/2) + i (rot3 / 2)) = pi / 3 + 2 pi * n, where n is ... -3,-2,-1,0,1,2,3...