Question

Can anyone please help me with the following:
Determine the fourth roots of -16 in the form x+iy
Thank you

Answer 1

-16 = 16* (-1)

You want the 4th roots (as in 4th roots of unity) or the fourth root (1+i)/sqrt2

Answer 2

The fourth roots of -16 in the form x+iy, please.

Answer 3

Put -1 = 1(cos pi + i sin pi) in polar and use de Moivre

Answer 4

U have -16 = 16 * (-1) so 4th root of 16 is 2 and then use de Moivre on the -1

Answer 5

So, consider some polar number \(n\) with coordinates \(r, \theta\), we then have, by Euler:
\[
n=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta=re^{i\theta}
\]I'll be using the latter formula as opposed to the other two, simply because it's easier to type.
So, by Euler's formula (or an extension to de Moivre's), \(n^\frac{1}{l}\) is:
\[
\sqrt[l]{n}=\sqrt[l]{re^{i\theta}}=\left(\sqrt[l]{r}\right)\left(e^{i\left(\frac{\theta+2k\pi}{l}\right)}\right)
\]For some integer \(k\). So, we apply this, to find:
\[
n=16e^{i\pi}
\]So:
\[
\sqrt[4]n=2e^{i\frac{\pi}{4}}=2\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)=\sqrt{2}+i\sqrt{2}\\
\sqrt[4]n=2e^{i\frac{\pi+2\pi}{4}}=2\left(\cos\left(\frac{\pi+2\pi}{4}\right)+i\sin\left(\frac{\pi+2\pi}{4}\right)\right)=-\sqrt{2}+i\sqrt{2}\\
\sqrt[4]n=2e^{i\frac{\pi+4\pi}{4}}=2\left(\cos\left(\frac{\pi+4\pi}{4}\right)+i\sin\left(\frac{\pi+4\pi}{4}\right)\right)=-\sqrt{2}-i\sqrt{2}\\
\sqrt[4]n=2e^{i\frac{\pi+6\pi}{4}}=2\left(\cos\left(\frac{\pi+6\pi}{4}\right)+i\sin\left(\frac{\pi+6\pi}{4}\right)\right)=\sqrt{2}-i\sqrt{2}\\
\]And those are the fourth roots of \(n\).

Answer 6

Can't thank you enough! I'm on my way to make sense of it just now for myself... Thank you!

Answer 7

Haha, sure thing