Find all solutions of the equation $4(\overline{z})^4+z^2=0$

Question

I need to find all of the solutions of:

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

$\overline{z}$ is the conjugation of z.

I know that $z=a+ib$ and $\overline{z}=a-ib$, then: $4(\overline{z})^4+z^2=4(-z)^4+z^2=4z^4+z^2=0$

and from here, I can factor it like so: $4z^4+z^2=z^2(4z^2+1)$ and then I can seperate it into two equations:

1. $z^2=0 \Leftrightarrow z=0$
2. $z^4-1=0 \Leftrightarrow z=-\frac{i}{2},z=\frac{i}{2}$.

But I'm still missing 3 more complex solutions (According to WolframAlpha):

1. $z = \frac{1}{4}(\sqrt(3) + -i)$
2. $z = \frac{1}{4}(-\sqrt(3) + i)$
3. $z = \frac{1}{4}(-\sqrt(3) + -i)$

And I'm unsure how to get them.

Also, somebody pointed it out to me that the way I did it is wrong, so I was thinking on substituting like so: $4(a-ib)^4+(a+ib)^2=0$, but it gives completely different result...

Answer 1

Let $z=re^{i\theta}$ as usual. Taking the modulus of both sides in the equation $4(\bar z)^4=-z^2$, we get $4r^4=r^2$. So, $r=0$ or $r=\frac12$. If $r=0$, $z=0$. But, $z=0$ is not interesting solution. So, $z=\frac12e^{i\theta}$ and hence we get $e^{-4\theta i}+e^{2\theta i}=0$ that is $e^{6\theta i}=-1$. So, $\theta=\frac\pi 6+\frac{k}{3}\pi$ where $k$ can be $0,1,2,3,4,5$. Hence all, $7$, solutions are $$\{0,\frac{\sqrt 3+i}{4},\frac{i}{2},\frac{-\sqrt 3+i}{4},\frac{-\sqrt 3-i}{4},-\frac{i}{2},\frac{\sqrt 3-i}{4}\}.$$

Answer 2

HINT

I would start with noticing that \begin{align*} 4(\overline{z})^{4} + z^{2} = 0 & \Rightarrow 4|z^{4}| = |z^{2}|\\\\ & \Rightarrow |z|^{2}(4|z|^{2} - 1) = 0\\\\ & \Rightarrow (|z| = 0)\vee(|z| = 1/2) \end{align*}

Since $z = 0$ is a solution, we can restrict our attention to the case where $|z| = 1/2$: \begin{align*} 4\left(\frac{e^{-i\theta}}{2}\right)^{4} + \left(\frac{e^{i\theta}}{2}\right)^{2} = 0 & \Longleftrightarrow e^{-4i\theta} + e^{2i\theta} = 0\\ & \Longleftrightarrow e^{6i\theta} + 1 = 0\\\\ & \Longleftrightarrow e^{6i\theta} = -1 \end{align*}

Can you take it from here?

EDIT

As an answer to the comments related to this solution, we shall prove that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

The exponential function $e^{x}$ can be defined in terms of power series as follows: \begin{align*} e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \ldots \end{align*}

which converges for every $x\in\mathbb{R}$ (you may apply the ratio test, for example).

Based on it, consider the expression for $e^{ix}$, which is given by \begin{align*} e^{ix} & = 1 + ix - \frac{x^{2}}{2!} - \frac{ix^{3}}{3!} + \frac{x^{4}}{4!} + \frac{ix^{5}}{5!} \ldots\\\\ & = \left(1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \ldots\right) + i\left(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \ldots\right)\\\\ & = \cos(x) + i\sin(x) \end{align*}

where such power series converges absolutely for every $x\in\mathbb{R}$ due to the ratio test, for example.

Having said that, we can formally substitute $e^{i\theta}$ by $\cos(\theta) + i\sin(\theta)$, where $\theta\in\mathbb{R}$.

Answer 3

Another way without using exponential form, by conjugation we obtain

$$4\overline{z}^4+z^2=0 \iff 4z^4+\overline{z}^2=0$$

and subtracting

$$4(\overline{z}^4-z^4)=\overline{z}^2-z^2 \iff \begin{cases} \overline{z}^2-z^2=0 \implies \overline{z}^2=z^2 \\\\ \overline{z}^2+z^2=\frac14 \implies \overline{z}^2=-z^2+\frac14 \end{cases}$$

and substituting in the conjugated original equation, we obtain

$$\begin{cases} 4z^4+z^2=0 \implies z=0,z=\frac{\pm i}{2}\\\\ 4z^4-z^2+\frac14=0 \implies \frac{\pm \sqrt 3\pm i}{4}\end{cases}$$