For which real values of α does the following equation have non-zero solutions? z + z* = α|z| . Sketch the set of those solutions for any such value of α.
I gained values of 2 and -2 for α, but realise I am unable to sketch this in the complex plane.
On
As you observed in the comments, if $z + z^* = \alpha\lvert z\rvert$ where $z=x+iy$ then
$$2x=a\sqrt{x^2+y^2}.\tag1$$
Notice that $x = y = 0$ is a solution regardless of the value of $\alpha.$ But if we consider the case where $x$ and $y$ are not both zero, Equation $(1)$ says that $x$ must have the same sign as $\alpha.$ That is, if $\alpha$ is positive $x$ then must be positive, if $\alpha$ is negative then $x$ must be negative, and if $\alpha = 0$ then $x = 0.$
Solving for $y$ in Equation $(1)$ in the case where $\alpha \neq 0$,
$$ y = \pm \left( \frac{\sqrt{4 - a^2}}{a} \right) x. \tag2$$
(Almost all the details of this solution are already worked out in another answer, so I will not repeat them.)
Equation $(2)$ gives you equations of two lines, but remember that we still have the constraint that $x$ must have the same sign as $\alpha$ (unless $x = y = 0$). So you get only a part of each of the lines.
Try sketching the solution set for $\alpha = \sqrt2.$ Then try $\alpha = -\sqrt2.$ Can you generalize?
Let $z=x+iy$ and rewrite the equation,
$$2x=a\sqrt{x^2+y^2}$$
$$4x^2= a^2 (x^2+y^2)$$
$$(4-a^2)x^2=a^2y^2$$
$$\sqrt{4-a^2}x=\pm ay$$
Thus, for $-2\le a \le 2$, there are non-zero solutions, which are in general two lines through origin in the complex plane for a given $a$.
Note that, for $0<a<2$, $x>0$ from $2x=a|z|$. So, the lines in the 1st and 2nd quadrants are the valid solutions. For $-2<a<0$, the lines are in the 3rd and 4th quadrants. The solutions degenerate to one line if $ a=0, \pm 2$.