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Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck.

Given a general quadratic equation $az^2 + bz + c = 0$ with $a \not= 0$.

Using the same algebraic manipulation as in the case of real coefficients, we obtain $$a\left[\left(z + \frac{b} {2a}\right)^2 - \frac{\Delta} {4a^2}\right] = 0$$

This is equivalent to $$\left(z + \frac{b} {2a}\right)^2 = \frac {\Delta} {4a^2}\quad\text{or}\quad \left({2az + b}\right)^2 = \Delta$$ where $\Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to $$y^2 = \Delta = u + vi$$ where $u$ and $v$ are real numbers.

What I don't understand now is how the author obtained $$y_{1,2} = \pm \left(\sqrt { \frac {r + u} {2} } + \operatorname{sgn} (v)\sqrt{ \frac{r - u} {2} }\,i\right)$$ where $r = |\Delta|$ and $\operatorname{sgn}(v)$ is the sign of the real number $v$.

Robert Z
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Icosahedron
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2 Answers2

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Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain $$s^2-t^2=u,\quad 2st=v.$$ Now we solve this system with respect to $s$ and $t$. From the second $t=v/(2s)$ and plugging it in the first we get $$s^2-\frac{v^2}{4s^2}=u \Leftrightarrow 4s^4-4us^2-v^2=0$$ and we obtain $$s^2=\frac{u+\sqrt{u^2+v^2}}{2}=\frac{|\Delta|+u}{2}$$ Note that we dropped the part with the minus sign because $s^2\geq 0$. Finally $$t^2=\frac{v^2}{4s^2}=\frac{v^2}{2(|\Delta|+u)}=\frac{v^2(|\Delta|-u)}{2(|\Delta|^2-u^2)}=\frac{|\Delta|-u}{2}.$$ The we take the square roots keeping in mind that $\mbox{sgn}(s\cdot t)=\mbox{sgn}(v).$

Robert Z
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  • I understand your solution and thanks for the response but what I don't understand is this. If $y = s + it$ and $y^2 = u + iv$ then $s^2 - t^2 = u$ and $2st = v$ how did you arrive at this system? Please do explain your steps. – Icosahedron Jun 26 '17 at 10:03
  • Thanks your your response @Robert – Icosahedron Jun 26 '17 at 10:08
  • @Icosahedron You are welcome. BTW if you are new here please take a few minutes for a tour https://math.stackexchange.com/tour – Robert Z Jun 26 '17 at 10:11
  • Oops, never mind. I finally figured it out. You equated the real and imaginary part. THANKS a bunch. – Icosahedron Jun 26 '17 at 10:11
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Discriminant $\Delta = b^2 - 4ac = \alpha + i\beta$

We need to find the square root of $\Delta$ $\sqrt{\alpha + i\beta} = p + iq$ $\implies \alpha + i\beta = p^2 - q^2 + 2ipq$ $\implies p^2 - q^2 = \alpha$ and $ 2pq = \beta$ $\implies p^2 + q^2 = \sqrt{\alpha^2 + \beta^2}$ $\implies p = \sqrt{\dfrac{\alpha + |\Delta|}{2}}$ and $ q = \sqrt{\dfrac{|\Delta| - \alpha}{2}}$

$\implies \sqrt{\Delta} = \sqrt{\dfrac{\alpha + |\Delta|}{2}} + \dfrac{\beta}{|\beta|} \sqrt{\dfrac{|\Delta| - \alpha}{2}}i$

Now you can use the quadratic formula.

Rahul B.
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