Solving a complex polynomial including a conjugate

Question

I'm trying to solve the following polynomial but have been finding it difficult. I've tried looking on here for similar problems but have not found similar ones. Any help would be much appreciated!

The equation is (sorry about the poor format but I can't post pictures, hopefully it makes sense):

$$z^2 + 3\bar{z} - 3z = 2$$

Answer 1

Hint

Let $z=a+ib$,$\bar z=a-ib$ Replace in the expression to get two equations for two variables $a$ and $b$.

Just do it; it is quite simple.

By the way, Welcome to the site !

Answer 2

$$\Re^2(z)-\Im^2(z)+2i\Re(z)\Im(z)-6i\Im(z)=2$$

shows that $$(\Re(z)-3)\Im(z)=0.$$

Hence the solutions

$$\Re(z)=3,\\\Im^2(z)=7$$ and

$$\Im(z)=0,\\\Re^2(z)=2.$$

Answer 3

Let $z=r(\cos t+i\sin t)$ where $t,r\ge0$ are real

$$2=r^2\cos2t+(r^2\sin2t-6r\sin t)i$$

Equating the imaginary parts

As $r\ne0,$ $$2\sin t(r\cos t-3)=0$$

If $\sin t=0,\cos t=?,\cos2t=1,$

Equate the real parts

$2=r^2\cos2t=r^2,r=+\sqrt2$

If $r\cos t-3=0,r=?$

$$2=r^2\cos2t=r^2\cdot (2(3/r)^2-1)$$