Consider the equation $$ az^2+bz+c=0, $$ where $a,b$ and $c$ are complex numbers and $a\ne 0$. Applying usual operations on $\mathbb{C}$ we have the following: $$ az^2+bz+c=0\iff z^2+\frac{b}{a}z = -\frac{c}{a}\iff z^2+\frac{b}{a}z + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} -\frac{4ac}{4a^2} \\ \iff \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. $$ Now here's where my doubt comes. I know I can take the square root on both sides but wouldn't I have two results on both sides? Moreover, on the denominator of the right side (and on the whole left side) I have $z_0^2$; taking square root of this means I have only one solution ($z_0$)? Because I was told that the solution for the quadratic equation will be the same as for real numbers.


What you've done is correct, right up to the end. Essentially you've rederived the "quadratic formula".

The last step isn't taking the square root of both sides, it's equating the quantity in parentheses on the left to one of the two square roots of the complex number on the right. That will give you the two solutions to the original equation.

That will be just one solution counted twice if the discriminant is $0$, and will recover the usual real roots of real quadratics if the coefficients are real and the discriminant nonzero.

  • $\begingroup$ Ok, but I can't say that the two solutions will be $-\frac{b}{2a}\pm(\frac{b^2-4ac}{4a^2})^{1/2}$, so how I write the two solutions to the same form of the real case? $\endgroup$ – AnalyticHarmony Sep 24 '17 at 13:15
  • $\begingroup$ Moreover, if I have $(4a^2)^{1/2}$ I just can't equal this to $2a$ because I have two square roots here. $\endgroup$ – AnalyticHarmony Sep 24 '17 at 13:21
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    $\begingroup$ Think of the expression on the right as a single complex number $u+iv$ after you do all the arithmetic there. If it's not zero it has two complex square roots. Don't think about "square rooting" the numerator and denominator separately. You should not use the exponent $1/2$ or the square root symbol for complex numbers. It leads to confusion. $\endgroup$ – Ethan Bolker Sep 24 '17 at 13:24

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