How to solve $z$ and 3 other roots when $z^{\frac{1}{4}}=2+i$

Question

If complex number $z$ has 4 roots and value of first root $w_0=2+i $, how can I solve 3 other roots and number $z$ (to $a+bi$)?

I know I could use this for solving roots:

$w_k=w_0*p^k$ where k is wanted root (1 for next root etc.) and $p=e^{i\frac{(2\pi)}{n}}$ ($n$ = total number of roots)

Problem is I do not know what is $w_0$ in same form as $p$ above or how to solve for it.

We have an example in our book (if this helps to understand my problem) where $z$ is solved for 3 roots like this:

$z=8e^{i(\frac{\pi}{2})}$ where $z^{\frac{1}{3}}= w_0= 2e^{i\frac{\pi}{6})} = \sqrt3+i$ and from here you can solve for $w_1$ and $w_2$ with $w_k=w_0*p^k => w_1 = 2e^{i(\frac{\pi}{6}+\frac{2\pi}{3})}= -\sqrt3 + i$

Answer 1

If $2+i$ is a fourth root of $z$, then the remaining fourth roots of $z$ are:

• $i\times(2+i)=-1+2i$;
• $(-1)\times(2+i)=-2-i$;
• $(-i)\times(2+i)=1-2i$

since the fourth roots of $1$ are $1$, $i$, $-1$, and $-i$:

Answer 2

To find the other fourth roots, given one of them, note that they are equally spaced around a circle of radius $|2+i|=\sqrt 5.$ Thus, since one of the roots is $2+i=\sqrt 5e^{\phi i},$ where $\tan\phi=1/2,$ the other roots are then $$\sqrt 5e^{(\phi+2πk/4) i},$$ with $k\in\{1,2,3\}.$