Derivation of the quadratic formula for complex coefficients

Question

I am aware that the formula $Az^2 + Bz + C = 0$ works for solving quadratic equations for complex numbers, whether the coefficients are complex or not. I need too prove that if $B,C$ are complex numbers; $B,C \in \mathbf{C} $, that the quadratic formula can be written like this:

$$ z_{n} = \frac{-B + \sqrt{|B^2 - 4C|} e^{in\pi} e^{\frac{i}{2}arg(\frac{1}{4}B^2-C)} }{2}; n \in {[0, 1]} . $$

My interpretation of the expression is this:

1) $\frac{-B + \sqrt{|B^2 - 4C|}}{2}$ equals the modulus of $Z_{n}$.

2) $e^{in\pi}$ equals the sign of the real number, and will be -1 or 1 depending on $n$. Multiplied with the modulus of $Z_{n}$, it will give the co-ordinate of the real number.

3) I know that the complex numbers in exponential form can be written as $e^{i \frac{\pi}{2}}$, so I believe that the expression $e^{\frac{i}{2}arg(\frac{1}{4}B^2-C)}$ has something to do with the complex part of the number.

My questions are connected to part 3):

1) Is $arg$ equal to $\pi$, or is it an arbitrary angle?

2) Why is $arg$ multiplied with $(\frac{1}{4}B^2-C)$ ?

3) How is the expression $e^{i \frac{\pi}{2}}$ related to $e^{\frac{i}{2}arg(\frac{1}{4}B^2-C)}$ ?

Answer 1

The equation $a z^2 + b z + c = 0$ rewrites as \begin{equation} \left(z + \frac{b}{2 a}\right)^2 = \frac{b^2 - 4 a c}{4 a^2}\, . \end{equation} If $b^2-4ac = 0$, then the solution is $z = -b/2a$. Else, we introduce the polar decomposition \begin{equation} \frac{b^2 - 4 a c}{4 a^2} = r e^{\text{i}\theta} \qquad\text{with}\qquad r = \left|\frac{b^2 - 4 a c}{4 a^2}\right| \quad\text{and}\quad \theta = \arg\left(\frac{b^2 - 4 a c}{4 a^2}\right) \, , \end{equation} of which $z+b/2a$ is a square root in $\mathbb{C}$. Therefore, \begin{equation} z_n = -\frac{b}{2 a} + \sqrt{r}\; e^{\text{i}\theta/2} e^{2\text{i}\pi n/2} \, , \qquad\text{where}\qquad n\in \lbrace 0, 1\rbrace \, . \end{equation} The formula in the question corresponds to the case $a=1$.