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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.
Find all the roots of z^{4}=1-i
I guess I should state De Moivre's here...
(r cis(\vartheta))^{n}=r^{n} cis (n\vartheta)
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})
Homework Statement
Find all the roots of z^{4}=1-i
Homework Equations
I guess I should state De Moivre's here...
(r cis(\vartheta))^{n}=r^{n} cis (n\vartheta)
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})