# Proving a simple equation with complex numbers

Fix $A \in ℂ$ and $B \in ℝ$ Let $z \in ℂ$.

Show that the equation $|z^2| + Re(Az) + B = 0$ has solutions iff $|A^2| ≥ 4B$

I have no trouble proving the forward implication, but its the "only if" part that I can't prove. I am writing the complex numbers in polar form and then working from there with the discriminant. As far as I can tell, the $|A^2| ≥ 4B$ is not a sufficient condition to imply that the equation will have solutions. Can someone help me to see what I am doing wrong?

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## 2 Answers

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

So, we have $$x^2+y^2+(xa-yb)+B=0\iff (2x+a)^2+(2y-b)^2=a^2+b^2-4B$$

As $x,y,a,b$ are real, the left hand side must be $\not<0$

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Suppose that $|A|^2\ge4B$. The equation can be written as $$4z\overline z+2Az+2\overline A\overline z=-4B$$ and hence as $$|2z+\overline A|^2=|A|^2-4B\ .$$ The RHS is a non-negative real number, say $r^2$, and so the equation has solutions. In fact, the complete solution is $$z=\frac{re^{i\theta}-\overline A}{2}$$

with $\theta$ real.

Comment. In fact, this is really just the same as @lab bhattacharjee's solution. But IMO it is a nice illustration of the simplification one can obtain by writing moduli and real parts (and imaginary parts if applicable) in terms of conjugates.

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