Cube roots

[BobbyJones]

I have a question that asks: Find the cube roots of 32(SQRT)2 + j 32(SQRT)2 in polar form.

I worked the numbers out in to decimals and it was about 45. something, I then found r by squareing each, adding them together then square-rooting that answer, came to 14 if I remember. As for the angle I took the inverse of tan(Y/x), came to 45degrees.

Am I correct to then cube root r and simply divide the angle by 3?


Before anyone asks (SQRT) means Square root. I just couldnt do the symbol.

 

[soroban]

Hello, BobbyJones!

\(\displaystyle \text{Find the cube roots of }z \:=\:32\sqrt{2}+ 32i\sqrt{2}\text{ in polar form.}\)

Am I correct to then cube root \(\displaystyle r\) and simply divide the angle by 3? . Well, yes and no.

Yes, you found one of the cube roots.

No, because they asked for all the cube roots.


In polar form, we have: .\(\displaystyle z \;=\;64\left[\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right]\)

Generalized, we have: .\(\displaystyle z \;=\;64\left[\cos\left(\frac{\pi}{4} + 2\pi n\right) + i\sin\left(\frac{\pi}{4} + 2\pi n\right)\right]\)


You are correct: cube root the \(\displaystyle r\), divide the angle by 3.

. . \(\displaystyle z^{\frac{1}{3}} \;=\;4\bigg[\cos\left(\frac{\pi}{12} + \frac{2\pi}{3}n\right) + i\sin\left(\frac{\pi}{12} + \frac{2\pi}{3}n\right)\bigg] \)


Now let \(\displaystyle n = 0,1,2\)