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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 $e^{\frac{i}{2}arg(\frac{1}{4}B^2-C)}$ ?

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  • $\begingroup$ $\arg$ is not a number.It is a function. If $0\ne z=a+ib$ with real $a,b$ then $\arg(z)$ satisfies $z=|z|e^{i\arg (z)},$ where $|z|=\sqrt {a^2+b^2}$......... $\arg(0)$ is not defined, but if $B^2=4C$ then $\sqrt {B^2/4-C}=0$ so we can assign an arbitrary value to $\arg(0)$ in the formula in your Q $\endgroup$ Mar 14, 2017 at 22:32
  • $\begingroup$ See another answer here $\endgroup$ Mar 15, 2017 at 3:33

1 Answer 1

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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 $a=1$.

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