Can a quadratic involving complex inputs be solved by the quadratic formula as well? If a quadratic equation $\gamma(z) = az^2 + bz + c = 0,$ where $a,b,c \in \mathbb{R},$ takes as input values from the complex plane, may one use the quadratic formula to solve the quadratic equation? Won't these solutions necessarily involve complex numbers whose $y$-components are $0$, since the quadratic formula would be some algebraic expression of real numbers? Thank you! 
$$az + b = cz^2 + dz \implies cz^2 + (d-a)z - b = 0$$
Applying the quadratic formula:
$$z = \frac{-(d-a) \pm \sqrt{(d-a)^2 + 4cb}}{2c}.$$
Would I leave the solution simply in this form?

For your particular problem, if $a, b, c,$ and $d$ are all real numbers, there are three possibilities:


*

*The expression under the square root sign, $(d-a)^2 + 4cb,$ is a positive real number. Then you have two solutions for $z,$ both of which have zero imaginary component.

*The expression under the square root sign, $(d-a)^2 + 4cb,$ is zero. Then you have one solution for $z$ (also called a double root) and it has zero imaginary component.

*The expression under the square root sign, $(d-a)^2 + 4cb,$ is a negative real number. Then you have two solutions for $z,$ both of which have non-zero imaginary component, and both of which have the same real component.
In fact, in the third case the two solutions for $z$ will be what are called complex conjugates: the real component is the same and the imaginary component of one is exactly opposite the other.
The sum of the two solutions has zero imaginary component,
so also does the arithmetic mean of the two solutions (which is just half the sum).
