Roots of $(z^2-z)^4=81$ in polar form

Question

Question :

Find the roots of $(z^2-z)^4=81$ in polar form.

What i've done :

I'm thinking about using De Moivre's Theorem.

Let $w=z^2-z \Rightarrow w^4=81$

$\begin{align} &\Leftrightarrow w^4=3^4\,\text{cis $2\pi$}\\ &\Leftrightarrow w=3\,\text{cis $\left(\dfrac{2\pi+2k\pi}{4}\right)$}\\ &k=0,1,2,3 \end{align}$

What's next? I'm confusing to substitute $w=z^2-z$ back...

Please help me. Thanks

You could use $z^2-z=(z-\frac12)^2-\frac14$. – kingW3 Oct 10 '19 at 00:14 — kingW3, Oct 10 '19 at 00:14
Is it completing the square? – user516076 Oct 10 '19 at 00:17 — user516076, Oct 10 '19 at 00:17
And apply the de moivre's once again? – user516076 Oct 10 '19 at 00:17 — user516076, Oct 10 '19 at 00:17
Yes, that's how to proceed. – Toby Mak Oct 10 '19 at 00:26 — Toby Mak, Oct 10 '19 at 00:26

score 2 · Answer 1 · answered Oct 10 '19 at 01:03

I like your idea that first, you denote $$w=z^2-z.$$ Then, recall that in polar form, you have $$c = r e^{i \theta} ,$$ where $c \in \mathbb{C}$ is any complex number, $r \in \mathbb{R}$ is its distance from the origin, and $\theta \in [0,2\pi)$ is its phase.

So now, you can first find the first root, i.e. $w_1$, in a "regular way": $$w_1 = \sqrt[4]{81} = 3.$$

Note that this is the root located on $x$ axis, so its angle $\theta$ in polar form is $0$, which is why $w_1 = r e^{i \theta} $ lets you determine the $r$ distance (since $e^{i \cdot 0} = 1$). Thus you have that $r=3$

Next, since you know that you are taking the fourth root, by the fundamental theorem of algebra, you know that you must have $4$ roots. Then, taking a shortcut, we have $\Delta \theta = 2 \pi /4 = \pi/2$. (Or you can also write it in a more formal way with De Moivre's theorem in polar form.) Then, your other three roots would be:

$$w_2 = re^{i \pi/2}=3 e^{i \pi/2} = ...$$

$$w_3 = re^{i 2\pi/2}=3 e^{i 2\pi/2} = ...$$

$$w_4 = re^{i 3\pi/2}=3 e^{i 3\pi/2} = ...$$

Compute these answers; for each of the cases then substitute into $ z^2-z -w=0$ and proceed to computing $z$ for each case (with the computation of the determinant and using De Moivre or the shortcut described above).

Ok, i kind of understand, will it gives me $8$ roots in total? Cz when we expand the $z$ it has $8$ orders? — user516076, Oct 10 '19 at 01:16
That is correct, 8 roots; however, they are not necessarily distinct according to the fundamental theorem of algebra (I would expect it to be 8 distinct but would be better to numerically check probably). — Elen Khachatryan, Oct 10 '19 at 01:19

Roots of $(z^2-z)^4=81$ in polar form

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