Complex Roots - Not sure I did this right

1. Mar 27, 2009

Atena

Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering.

1. The problem statement, all variables and given/known data

Find all the roots of $$z^{4}=1-i$$

2. Relevant equations

I guess I should state De Moivre's here...

$$(r cis(\vartheta))^{n}=r^{n} cis (n\vartheta)$$

3. The attempt at a solution

Firstly I re-wrote $$z^{4}=1-i$$ as

$$z^{4}=\sqrt{2} cis (\frac{-\pi}{4})$$

Using De Moivre's,

$$z=(2\frac{1}{2})^{\frac{1}{4}} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))$$

$$z=2\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+k2\pi))$$

I found the four roots letting k=0,1,2,3

$$z=2^\frac{1}{8} cis (\frac{-\pi}{16})$$

$$z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+2\pi))=2^\frac{1}{8} cis (\frac{7\pi}{16})$$

$$z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+4\pi))=2^\frac{1}{8} cis (\frac{15\pi}{16})$$

$$z=2^\frac{1}{8} cis (\frac{1}{4}(\frac{-\pi}{4}+6\pi))=2^\frac{1}{8} cis (\frac{23\pi}{16})$$

2. Mar 27, 2009

yyat

Hi Atena!

Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).