Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$.

Question

Determine all six exact values if $\sqrt[6]{\frac{2i}{1+i}}$, expressing each in simplified, exact rectangular form.

What I have done: I set $\sqrt[6]{\frac{2i}{1+i}}=z$ and put the both to the sixth power to get ${\frac{2i}{1+i}}=z^6$. I then multiplied by the conjugate to get $z^6(1-i)=\frac{2i-2i^2}{2}$ and simplified to $z^6(1-i)=i-i^2$ and simplified farther to get $z^6-z^6i=i+1$. I then moved the equation to one side to set it equal to zero and got $z^6-z^6i-i-1=0$.

This is where I got stuck, and I do not know if I even did this right.

Can anyone help me, please?

Answer 1

Use polar coordinates and De Moivre Theorem

Let $w = \frac {2i}{1+i}=\frac {2i(1-i)}{(1+i)(1-i)} =\frac {2+ 2i}{2} = 1 + i$

$w = re^{i(\theta + 2\pi k)} = r(\cos \theta + i\sin \theta) = 1 + i$ where

$r = \sqrt {1^2 + 1^2} =\sqrt 2$ and $\theta = \arctan \frac 11 = \frac \pi 4$.

So $w = \sqrt 2*e^{i(\frac \pi 4+ 2\pi k)}$.

So $z = w^{\frac 16} = \sqrt[6]{\sqrt 2}e^{i(\frac {\frac \pi 4+ 2\pi k}6)}$ where $k=0,...5$.

$= \sqrt[12]{2} (\cos \phi + i \sin \phi)$ where $\phi = \frac \pi {24}+ \frac 13\pi k$ where $k = 0... 5$ are the six sixth roots.

Answer 2

You multiplied both sides with the conjugate, what you should do is multiply the $\frac{2i}{i+1} $ by $\frac {1-i}{1-i} $ ($=1$) to get rid of the $i$ in the denominator.

Answer 3

Note that

$$z^6=\frac{2i}{1+i} =\frac{2e^{i \frac\pi2}}{\sqrt2e^{i \frac\pi4}} =\sqrt2e^{i \frac\pi4}$$

Then,

$$z= 2^{\frac1{12}} e^{\frac {i}6(\frac\pi4+2n\pi)} = 2^{\frac1{12}} e^{i\pi(\frac1{24}+\frac n3)}$$

with $n=0,1,2,3,4,5$.