Fourth roots of $-17$

The question I'm stuck on is as follows:

Find all $4$th roots of $−17$ in Cartesian form. Simplify as much as possible.

Here's what I've done so far:

$$z^4 = -17\\ |z| = \sqrt{(-17)^2} = 17 = r\\$$Using De Moivre's Theorem:$$-17 = 17e^{0i +2kπi}\\ -17^{\frac{1}{4}} =17^\frac{1}{4}[e^{2kπi}]^\frac{1}{4}\\ 17^\frac{1}{4}[e^{2k}]^{\frac{π}{4}i}\\ z_0 = 17^\frac{1}{4}[\cos(0)+\sin(0)i] = 17^\frac{1}{4}\\ z_1 = 17^\frac{1}{4}[\cos\left(\frac{π}{2}\right) + \sin\left(\frac{π}{2}\right)i] = 17^\frac{1}{4}$$

• $e^{\pi i}=-1$ might help, although it might be easier to think of introducing the fourth roots of $i$. – Chris Leary Jan 30 '14 at 4:40

The step you have done$|z|$=17 is wrong.As $z^4$ = -17,hence $z^2$ = $\pm$ $\sqrt17$$\timesi. Case 1:- For z^2 = +\sqrt17$$\times$i.On square rooting i you will get $\sqrt i$ = $\frac{1}{\sqrt2}$$\times(1 + i). hence as z^2 = +\sqrt17$$\times$i so z=$\pm$ $\mathrm {17}^{1/4}$$\times$$\frac{1}{\sqrt2}$$\times(1 + i). Case 2:- For z^2 = -\sqrt17$$\times$i=$\sqrt17$$\times(-i).On square rooting -i you will get \sqrt -i = \frac{1}{\sqrt2}$$\times$(1 - i).Hence z=$\pm$ $\mathrm {17}^{1/4}$$\times$$\frac{1}{\sqrt2}$$\times(1 - i).Therefore z=\pm \mathrm {17}^{1/4}$$\times$$\frac{1}{\sqrt2}$$\times$(1 $\pm$ i)
• $z^2=\pm i \sqrt{17}$. $i$ has two square roots, the one you give and its negative. You should find four fourth roots of $-17$ – Ross Millikan Jan 30 '14 at 4:53
As $\displaystyle -1=e^{i\pi }=e^{(2k+1)\pi i}$ where $k$ is any integer
So, $\displaystyle -1^{\frac14}=e^{\frac{(2k+1)\pi i}4}$ where $k$ can assume any four in-congruent values $\pmod4$
Now use Euler Formula to get $\displaystyle -1^{\frac14}=\cos\dfrac{(2k+1)\pi}4+i\sin\dfrac{(2k+1)\pi}4$ where we can take $k$ to be $0,1,2,3$