Show complex solutions exist

Question

Let A be a complex number and B a real number. Show that the equation $\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point.

Well i am trying to do the first part first. So assuming the equation has a solution that would mean $z = x+iy$ satisfies the equation.

I was going to let $A = s+it$ for a complex number, but it is not working out for me. Wrong step?

Answer 1

We have $$ 0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B $$ If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B>0$, and hence no solutions.

If $4B>\lvert A\rvert^2\le 0$, set $C=\frac{1}{2}\sqrt{\lvert A\rvert^2-4B}$, and our equation is equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert^2=C^2, $$ and hence equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert=C, $$ the set of solutions of which is the circle centered at $-A/2$ with radius $C$.

Answer 2

You are on the right track by letting $z = x+iy$ and $A=s+it$. If substitute this in, we have $$ x^2+y^2+xs - yt+r = 0 $$ where $r$ is my real number. Try completing the square for $x$ and $y$ and see what you get.

Answer 3

Let $z=x+iy, A=a+ib$ where $x,y,a,b$ are real

So, we have $x^2+y^2+ax-by+B=0$

$\implies\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2=B-\dfrac{a^2+b^2}4$

But $\left(x+\dfrac a2\right)^2+\left(y-\dfrac b2\right)^2\ge0$

Hope you can take it from here