It be $a\in\Bbb C , b ∈\Bbb R$ and $|a|^2 > b$. Also, $a'$ is the conjugation of $a: a' = x - iy$ when $a = x + iy$ (and equally for $z$).
It needs to be shown, that the solutions of the equation:
$z z' − a z' − a z' + b = 0$
draw a circle in $\Bbb C$. Also, the center and radius of the circle need to be calculated.
Now, I've approached the problem by writing $a$ as $x + iy$ and $z$ as $z = v + iw$ (and equally for $z'$ and $a'$), which resulted in:
$(v + iw)(v - iw) − (x + iy)(v - iw) − (x - iy)(v + iw) + b = 0$
This equation is equivalent to:
$0 = v^2 + w^2 - 2(xy + yw) + b$
However, from this point on, I don't know what exactly to do in order to prove the statement. My idea was to somehow work with the absolute value of $z$, as the absolute value might somehow fit the radius of the circle (the distance from the center of the circle to all the points on the circle is always the same), but I didn't really have any idea on how to do so.
Thanks in advance for any help!
Note that $$ (z-a)\overline{(z-a)} = z\overline z - z\overline a - a\overline z + a\overline a. $$ Moreover, we have $a\overline a = |a|^2$. So we get $$ z\overline z - z\overline a - a\overline z + b = (z-a)\overline{(z-a)} + b - |a|^2. $$ From this, we see that $z\overline z - z\overline a - a\overline z + b = 0$ iff $$ (z-a)\overline{(z-a)} = |a|^2 - b. $$ By assumption we have $|a|^2 - b > 0$. Moreover, $(z-a)\overline{(z-a)} = |z-a|^2.$ So we get $$ z\overline z - z\overline a - a\overline z + b = 0 \quad iff \quad |z-a| = \sqrt{|a|^2 - b}.$$ It's easy to see that the solution set of the r.h.s. of the preceding equivalence is exactly the circle with center $a$ and radius $\sqrt{|a|^2 - b}.$