Question

Let z= (√3)-i. Express z^5 in both polar form and in the form a+bi.

Answer 1

you need first \(r=\sqrt{a^2+b^2}\) and then \(\theta\)
\[r=\sqrt{\sqrt{3}^2+1^2}=\sqrt{3+1}=2\]
\[\theta =\arctan(\frac{-1}{\sqrt{3}})=-\frac{\pi}{6}\] so in polar form it is
\[2(\cos(-\frac{\pi}{6})+i\sin(\frac{\pi}{6}))=2e^{-i\frac{\pi}{6}}\]

Answer 2

\[2(\cos(-\frac{\pi}{6})+i\sin(-\frac{\pi}{6}))=2e^{-i\frac{\pi}{6}}\] typo on that last line

Answer 3

then to raise to the fifth power, you need only raise \(2^5=32\) and multiply the angles by 5

Answer 4

you got it from there? clear or no?

Answer 5

Makes sense:) thanks! i'm a little confused on how to put it into a + bi form though, I'm having trouble with the radical...

Answer 6

ok it would be much easier of you drew a picture

Answer 7

but as far as the \(a+bi\) form is conserned, you only need to evaluate sine and cosine at
\[-\frac{5\pi}{6}\]

Answer 8

concerned *

Answer 9

you get
\[32(\cos(\frac{-5\pi}{6})+i\sin(\frac{-5\pi}{6}))\]
\[32(-\frac{\sqrt{3}}{2}+i(-\frac{1}{2})\]
\[-16\sqrt{3}-16i\]

Answer 10

thanks!!

Answer 11

yw