Question

Please help| Find the fourth roots of the complex number below.

Answer 1

\[\left( \frac{ 1+i }{ 2-i }\right)^{\frac{ 1 }{ 4 }}\]

Answer 2

First rewrite the given complex number:
\[\frac{1+i}{2-i}\times\frac{2+i}{2+i}=\frac{1}{5}+i\frac{1}{5}\]
or a different way that you might find more convenient:
\[\begin{cases}1+i=\sqrt2e^{i\pi/4}\\
2-i=\sqrt{5}e^{i\arctan(-1/2)}\end{cases}~~\implies~~\begin{align*}\frac{1+i}{2-i}&=\sqrt{\frac{2}{5}}e^{i(\pi/4-\arctan(-1/2))}\\&=\sqrt{\frac{2}{5}}e^{i(\pi/4+\arctan(1/2))}\end{align*}\]

Answer 3

Now, the \(n\)th roots to any complex number are given by
\[z^{1/n}=r^{1/n}\exp\left(i\frac{\theta+2\pi k}{n}\right)\]
where in this case, \(n=4\), and \(k=0,1,2,3\).

Answer 4

As an example, one fourth root is
\[\left(\sqrt{\frac{2}{5}}\right)^{1/4}\exp\left(i\frac{\frac{\pi}{4}+\arctan\frac{1}{2}}{4}\right)\]
which you can try to manipulate if you like. You can write \(\arctan\dfrac{1}{2}=\text{arccot}2\) and \(\left(\sqrt{\dfrac{2}{5}}\right)^{1/4}=\sqrt[\Large8]{\dfrac{2}{5}}\), for instance.

Answer 5

Slight mistake, the reduced form of the initial complex number is actually \(\dfrac{1}{5}+i\dfrac{\color{red}3}{5}\), so all you need to do is account for that error.

Answer 6

Ah ok that makes sense! thank you!!