Question

check my complex arithmetic?

evaluate: final result in polar form "*" means conjugate

((-.5 + i(sqrt(3)/2))*)^8 =
=(-.5 - i(sqrt(3)/2)^8
=(e^(i*2pi/3))^8
=e^(8 + i*2pi/3)

is this correct?

Answer 1

\[z^8=\left(-\frac{1}{2}+i\frac{\sqrt3}{2}\right)^8~~?\]

Answer 2

I just found the equation button 1 sec

Answer 3

I just have \[z = ((-\frac{ 1 }{ 2 } + i \frac{ \sqrt3 }{ 2 })^*)^8\]
where * means the conjugate

Answer 4

then I get
\[= (-\frac{ 1 }{ 2 } - i \frac{ \sqrt3 }{ 2 })^8\]
\[=(e ^{i \frac{ 2\pi }{ 3 }})^8\]
\[=e ^{8+i \frac{ 2\pi }{ 3 }}\]

Answer 5

Oh I see.
\[\begin{align*}\left(\left(-\frac{1}{2}+i\frac{\sqrt3}{2}\right)^*\right)^8&=\left(-\frac{1}{2}-i\frac{\sqrt3}{2}\right)^8\\\\
&=\left(-\frac{1}{2}-i\frac{\sqrt3}{2}\right)^8\\\\
&=\left(\frac{1}{2}+i\frac{\sqrt3}{2}\right)^8\\\\
&=\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)^8\\\\
&=\cos\frac{8\pi}{3}+i\sin\frac{8\pi}{3}&\text{(by DeMoivre's theorem)}\\\\
&=e^\cdots
\end{align*}\]

Answer 6

so in your 3rd line you can just get rid of the negatives? and they become positive?

Answer 7

I dont see where they went?

Answer 8

It was a manner of factoring. For example,
\[(-a-b)^2=\bigg((-1)(a+b)\bigg)^2=(-1)^2(a+b)^2=(a+b)^2\]
This is true for any even power.

Answer 9

oh ok great thanks! I understand now. you are very good at explaining and answering :)

Answer 10

You're welcome! And thank you!