Root of a complex number

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.

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malawi_glenn
Homework Helper
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.
Have you tried De Moivre's formula ?

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

malawi_glenn
Homework Helper
I believe the complex root is defined by (that is: it's usually continued from the real function by)
$$\sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2}$$
yes, and from De Moivre's formula we get the general:

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

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

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

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

malawi_glenn