# Math Help - nth roots of complex number fun!

1. ## nth roots of complex number fun!

I have been asked to find all the solutios for the following equation

$z^4 = 128(-\sqrt{3} + i)$

and leave my answers in exponential form and then plot them on an argand diagram.

I think I could probably do it if the 128 wasn't there. If it isn't there I put
$(-\sqrt{3} + i)$ on an argand diagram and find that the argument or radius from the centre is 2. I think I can then change that to
$(-\sqrt{3} + i)^\frac{1}{4}$ becoming $2e^{i\theta}$ or something along those lines.

Can anybody tell me how to put the 128 in and if I got the rest of it sort of correct? or if it is a trainwreck?

2. Originally Posted by Beard
I have been asked to find all the solutios for the following equation

$z^4 = 128(-\sqrt{3} + i)$

and leave my answers in exponential form and then plot them on an argand diagram.

I think I could probably do it if the 128 wasn't there. If it isn't there I put
$(-\sqrt{3} + i)$ on an argand diagram and find that the argument or radius from the centre is 2. I think I can then change that to
$(-\sqrt{3} + i)^\frac{1}{4}$ becoming $2e^{i\theta}$ or something along those lines.

Can anybody tell me how to put the 128 in and if I got the rest of it sort of correct? or if it is a trainwreck?
If $a+ bi= r e^{i\theta}$, then $128(a+ bi)= (128r)e^{i\theta)}$! What ever "r" you got for $-\sqrt{3}+ i$, multiply it by 128.

It happens that $128= 2^7$ so that $(128r)^{1/4}= 2^{7/4}r^{1/4}= (2)(2^{3/4})(r^{1/4})$.

3. Ahhhhhh, yes that makes sense, so, because my r = 2 I multiply that by 2^7 to get 2^8 and then the fourth root of that is 2^2 which is 4.

Thanks

4. Okay, so having done that step

I end up doing

$z^4 = 2^8e^{i\theta}$
and then $z = 4e^{\frac{i\theta}{4}}$ and then from there I just...

5. Originally Posted by Beard
Okay, so having done that step

I end up doing

$z^4 = 2^8e^{i\theta}$
and then $z = 4e^{\frac{i\theta}{4}}$ and then from there I just...

You have to find the argument $\theta$ , of course! $\theta=\arg(-\sqrt{3}+i)=\arctan\left(-\frac{1}{\sqrt{3}}\right)=\frac{5\pi}{6}$ , since $\cos\theta<0\,,\,\,\sin\theta>0$

Tonio