Finding all complex solutions $z$ of an equation involving $z$ and its conjugate $\bar z$

Question

I need to find all complex solutions of the equation: $2z + 2i\bar z = 0.$

This is what I have done so far:

Let $z = x + yi$ and $\bar z = x - yi$ and then substitute into the equation.

$2(x + yi) + 2i(x - yi) = 0$

$2x + 2yi + 2xi - 2yi^2 = 0$

$2x + 2yi + 2xi - 2y(-1) = 0$

$2x + 2yi + 2xi + 2y = 0$

$2x + 2y + 2yi + 2xi = 0$

$2(x + y) + 2(x + y)i = 0$

So now we setup the two equations which would be the same

$2(x + y) = 0$

$2(x + y) = 0$

When I checked the solutions to this on Wolfram Alpha it simply said $y = -x. $ Have I gone about this question the wrong way? What is the best way to approach this question and what are the correct solutions?

Answer 1

A geometric interpretation:

$$z+i \overline{z}=0.$$

The complex conjugate of $z$, $\overline{z}$, is found by reflecting $z$ by the real axis. Multiplying by $i$ is rotating by $\frac{\pi}{2}$ counterclockwise ($90^\circ$).

Take any point $z=(\alpha,-\alpha)$ on the line $y=-x$. Reflect it by the real axis and rotate by $90^\circ$ counterclockwise. You will get $-z=(-\alpha,\alpha)$ and when you add this to $z$, the result is zero.

Answer 2

Your answer $2(x+y)=0\iff x+y=0$ is the same as Wolfram Alpha's $y=-x$.

You could have said $2(x+y)+2(x+y)i=0\iff 2(x+y)(1+i)=0\iff x+y=0$.