# Roots of complex numbers

• Jul 29th 2008, 10:56 AM
dashreeve
Roots of complex numbers
I am just wanting to see if the book may have made a typo. When asking for the three cubed roots of z=-2+2i, we get sqrt(8) for r. the next step shows 6th root of 8 *((cos 135 + 360k)/(3) + i sin (135 + 360k)/(3)). I understand how to get everything here except for the 6??? Shouldn't nth root of r be 3rd root of sqrt (8)????
• Jul 29th 2008, 11:24 AM
wingless
Yes, and 3rd root of $\sqrt{8}$ is 6th root of 8.

$\sqrt[3]{\sqrt{8}}$

$(8^{\frac{1}{2}})^{\frac{1}{3}}$

$8^{\frac{1}{2}\cdot \frac{1}{3}} = 8^{\frac{1}{6}} = \sqrt[6]{8}$
• Jul 29th 2008, 11:42 AM
dashreeve
I see now, thank you for the help.