Question

Use De Moivre's theorem to write the complex number in trigonometric form.

Answer 1

\[(\sqrt2+i \sqrt2)^3\]

Answer 2

@mathstudent55

Answer 3

@triciaal do u know how to do this?

Answer 4

sorry don't remember the theorem

Answer 5

do u know someone that knows it?

Answer 6

or someone who might know?

Answer 7

@StudyGurl14 can you help?

Answer 8

he is offline :(

Answer 9

\[z=\left(\sqrt2+i\sqrt2\right)^3=\left(\sqrt2\right)^3(1+i)^3\]Write \(1+i\) in polar form:
\[1+i=\sqrt2e^{i\pi/4}\](because the point \((1,i)\) in the complex plane forms a right triangle with legs of length \(1\))

So you have
\[z=\left(\sqrt2\right)^3\left(\sqrt2e^{i\pi/4}\right)=\left(\sqrt2\right)^4\left(e^{i\pi/4}\right)^3\]DeMoivre's theorem states that
\[\left(\cos t+i\sin t\right)^n=\left(e^{it}\right)^n=e^{int}=\cos nt+i\sin nt\]which you can use to simplify the exponential factor. And of course \(\left(\sqrt2\right)^4=2^2=4\).

Answer 10

thanks, so the answer is \[(\sqrt2)^4=2^2=4\]

Answer 11

No, it's not. Partially because I made a mistake:
\[z=(\sqrt2)^3\left(\sqrt2e^{i\pi/4}\right)^\color{red}3=(\sqrt2)^6\left(e^{i\pi/4}\right)^3\]But the exponential term does *not* disappear.

Answer 12

ya cuz that is not one of my options

Answer 13

4i -4

Answer 14

Well, presumably you know that
\[e^{it}=\cos t+i\sin t\]right?

This means
\[e^{i\pi/4}=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1+i}{\sqrt2}\]DeMoivre's theorem says that
\[\left(e^{i\pi/4}\right)^3=e^{i3\pi/4}=\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}=\frac{-1+i}{\sqrt2}\]So you have
\[z=(\sqrt2)^6\left(\frac{-1+i}{\sqrt2}\right)=(\sqrt2)^5(-1+i)\]

Answer 15

sorry i had to do something, so now what do i do?

Answer 16

@HolsterEmission

Answer 17

the answer is: \[8(\cos(\frac{ 3\pi }{ 4})+i \sin (\frac{ 3\pi }{ 4}))\]