# Finding the four fourth roots of $-i$ using polar form

To start this problem I turned the complex number from Cartesian form to polar form by writing $$z^4=-i=\cos \left(\frac{3\pi}{2}+2\pi k \right)+i\sin \left( \frac{3\pi}{2}+2\pi k \right)$$ Where $k$ is equal to all integers. Once the equation for the roots has been written out k is substitued by any 4 consecutive integers in order to find the roots.

Next: $$z=\cos \left( \frac{3\pi}{8}+\frac{2\pi k}{4} \right)+i\sin \left( \frac{3\pi}{8}+\frac{2\pi k}{4} \right)$$ Thus we find that the four roots of $z^4=-i$ are $$z=\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}, \cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8}, \cos\frac{19\pi}{8}+i\sin\frac{19\pi}{8}, \cos\frac{3\pi}{27}+i\sin\frac{27\pi}{8}$$ My answer bookelt says that its wrong. Any ideas?

Thank :)

• Yo are making some msitakes in your calculations. For example, $\frac {3 \pi} 8+\frac {2\pi} 4=\frac {7 \pi} 8$. – Kavi Rama Murthy Aug 20 '18 at 9:36

## 1 Answer

You are almost right , I guess may be there's a mistake as you can put k =0,1,2&3. For k=0 your answer seems correct. Then for k=1 you will get 7π/ 8 as the angle. Similarly for k=2 &3 you yave angle as 11π/8 and 15π/8. This should be your final answer.