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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:$$ -17 = 17e^{0i +2kπi}\\ -17^{\frac{1}{4}} =17^\frac{1}{4}[e^{2kπi}]^\frac{1}{4}\\ 17^\frac{1}{4}[e^{2k}]^{\frac{π}{4}i}\\ z_0 = 17^\frac{1}{4}[\cos(0)+\sin(0)i] = 17^\frac{1}{4}\\ z_1 = 17^\frac{1}{4}[\cos\left(\frac{π}{2}\right) + \sin\left(\frac{π}{2}\right)i] = 17^\frac{1}{4} $$

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  • $\begingroup$ $e^{\pi i}=-1$ might help, although it might be easier to think of introducing the fourth roots of $i$. $\endgroup$ – Chris Leary Jan 30 '14 at 4:40
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The step you have done$|z|$=17 is wrong.As $z^4$ = -17,hence $z^2$ = $\pm$ $\sqrt17$$\times$i. Case 1:- For $z^2$ = +$\sqrt17$$\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)

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  • $\begingroup$ $z^2=\pm i \sqrt{17}$. $i$ has two square roots, the one you give and its negative. You should find four fourth roots of $-17$ $\endgroup$ – Ross Millikan Jan 30 '14 at 4:53
  • $\begingroup$ I corrected it but due to internet connection problem t didn't work out $\endgroup$ – Devgeet Patel Jan 30 '14 at 5:16
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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 $0,1,2,3$

de Moivre's formula should bring us in the destination

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