How to find results for $-z + \frac {3} {\overline {z}}=2$

Question

Can you just please help me solve this problem, because i don't know how to solve it: $$-z + \frac {3} {\overline z}=2$$

Answer 1

Hint: Multiplying both sides by $\bar z$ and using $z\bar z = |z|^2$, you will see that $2\bar z$ is a real number. So $\bar z = z$ and it becomes just a quadratic equation.

Answer 2

Multiply by $\bar{z}$ to get $$\vert z \vert^2 + 2z^* - 3 = 0$$ Let $z = x + jy$, so $$x^2 + y^2 + 2(x-jy) - 3 =0$$ which gives real part equal zero $$x^2 + y^2 +2x - 3 = 0$$ and imaginary part is zero $$-2y = 0$$ Hence $y = 0$ and $$x^2 + 2x - 3 = 0$$ Solve the above in $\mathbb{R}$ to get $x = -3$ and $x = 1$. So $z = -3$ and $z = 1$

Answer 3

We have that $z\not=0$ and by multiplying both sides by $\overline{z}$ we obtain $$-z\cdot\overline {z}+3=2\overline{z},$$ Now let $z=x+iy$ with $x,y\in\mathbb{R}$ and we get $$-(x+iy)(x-iy)+3=2(x-iy).$$ Can you take it from here?

Answer 4

$$-z + \frac {3} {\overline z}=2$$

$$-z \overline z+3=2\overline z$$

As $z \overline{z}=\left|{z^2}\right|$,

$$ \begin{align} -\left|{z^2}\right|+3&=2 \overline{z}\\ &= 2a-2bi \\ \end{align} $$

Equating imaginary parts,

$$2\Im{(z)}=0\implies 2b=0\implies b=0$$

Equating real parts,

$$ \begin{align} 2 \Re{(z)} &= -(a^2+b^2)+3 \\ 2a &= -a^2+3 \\ a^2+2a-3 &= 0 \\ (a+3)(a-1) &= 0 \\ a &= -3,\,1 \\ \end{align} $$

So $z=-3$ or $z=1$.

Answer 5

Multiply both sides by $\overline z$ and since $z\overline z=|z|^2$, you'll get $$ 2\overline z +|z|^2=3 $$ put then $z=a+ib$, from which last equation turns into $$ 2(a-ib)+a^2+b^2-3=0 $$

so you get $b=0$ and $a^2+2a-3=0$ once solved gives $a=1,-3$.

Thus the solutions to your equation are $z=1$ and $z=-3$.