Compare two complex numbers of trigonometric form in solving $z^4-16\bar{z}^2 = 0$

Question

I want to solve this:

$$z^4-16\bar{z}^2 = 0$$

So I represented both $z$ and $\bar{z}$ trigonometricaly and got:

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That's what I did, and my biggest worries is about how in the end to compare the $~\theta~$? I add $~2 \pi k~$ to one of $~\theta~$, is it the right way? should have I done it in other way?

$(z^2 - 4\bar{z})(z^2 + 4\bar{z}) = 0$

if $(z^2 - 4z) = 0 $

$\Rightarrow z^2 = 4z$

$r^2~\cos2\theta = 4r~\cos(-\theta)$

$*r^2 = 4r \Rightarrow r=0,4 , if r=0 \Rightarrow {z = 0}$

$** 2\theta = -\theta +2\pi k \Rightarrow \theta = (2/3)\pi k~~ $ is this line right? why or why not?

I know that I should proceed and do the same for $~z^2 + 4\bar{z}~$ but for now its true?

Answer 1

Set $z=re^{it}$ where $r\ge0$

$$r^4e^{i4t}=16r^2e^{-2it}$$

Taking modulus $r^4=16r^2\implies r=0, r^2=16$

$$e^{i6t}=1\implies6t=2m\pi$$ where $m$ is an integer

$t=\dfrac{2m\pi}6, t=0,1,2,3,4,5$

Answer 2

$$z^4=16.\bar z^2\,\,---(1)$$ Taking conjugate on both sides $$\bar z^4=16.z^2\,\,---(2)$$ from (1) $\bar z^2=\frac{z^4}{16}$ put in (2) $$\therefore \frac{z^8}{16^2}=16.z^2$$ $$z^2(z^6-16^3)=0$$ $$z=0 ; (\frac{z^2}{16})^3=1$$ $$z^2=0 , 16 , 16\omega , 16\omega^2$$ $$z=0,\pm 4 ,\pm 4\omega , \pm 4\omega^2$$

Answer 3

Multiply the given $z^4=16\bar{z}^2$ by its conjugate $\bar{z}^4=16{z}^2$ to get

$$|z|^8=16^2|z|^4$$

which leads to the trivial solution $z=0$ and $ |z|=4$. Then multiple $z^2$ to $z^4=16\bar{z}^2$

$$z^6=16\bar{z}^2z^2=16|z|^4=4^6$$

Thus, the non-zero solutions are $$z=\left(4^6e^{i 2\pi n}\right)^{1/6} =4e^{\frac{i \pi n}3}$$

with $n=0,...,5$

Answer 4

Why not put in trig terms first?

If $z = re^{\theta i}$ then

$r^4e^{4\theta i} = 16r^2 e^{-2\theta i}$

So $r^4 =16r^2$ and $4\theta = -2\theta + 2k\pi$ or $r^4 = 16r^2 = 0$

$r^2 = 16$ and $6\theta = 2k\pi$ or $r= 0$

$r = 4$ and $\theta = \frac k3\pi; k = 0.... 5$ or $r=0$.

As to your question:

Yes, your line is right. If $z^2 - 4\overline z$ then either $r = 0$ or $r=4$ and $\theta = \frac {2k}3\pi; k= 0,1,2$.

And if $z^2 +4\overline z=0$ then $z^2 = -4\overline z$ so $r = 0$ or $r = 4$ and then $\arg(-4\overline z) = \arg(-\overline z) = \arg(\overline z) + \pi = -\theta + \pi$ so $2\theta = -\theta + \pi + 2k\pi$ so

$3\theta = (2k+1)\pi$ and $\theta = \frac {2k+1}3\pi$ for $k=0; 1,2$.