What is the solution of this quadratic function

Question

$$iz^2 − 2 \sqrt{2}z − 2 \sqrt{3} = 0,\qquad z \in \mathbb{C}$$

I'm new to this site, i'm really thankful for your help

Answer 1

$$iz^2-2\sqrt{2}z-2\sqrt{3}=0\Longleftrightarrow$$ $$z=\frac{-\left(-2\sqrt{2}\right)\pm\sqrt{\left(-2\sqrt{2}\right)^2-4\cdot i\cdot\left(-2\sqrt{3}\right)}}{2\cdot i}$$

Simplify it and you'll get the roots!

Answer 2

You calculate as for any quadratic equation. First simplify it multiplying it by $-\mathrm i$, and get $$z^2+2\sqrt2\mathrm i z+2\sqrt3\mathrm i=0.$$ Now the reduced discriminant is $$\Delta'=(\sqrt2\mathrm i)^2-2\sqrt3\mathrm i=-2(1+\mathrm i\sqrt3)=4\mathrm e^{-\tfrac{2\mathrm i\pi}{3}},\quad\text{whence}\quad\sqrt{\Delta'}=\pm2e^{-\tfrac{\mathrm i\pi}{3}},$$ and the resolution formulae yield $$z=-\sqrt2\mathrm i \pm2e^{-\tfrac{\mathrm i\pi}{3}}.$$ There remains to simplify the results.

Answer 3

Ill continue further on Jan's part.$z=\frac{2\sqrt{2}\pm\sqrt{8+8\sqrt{3}i}}{2i}$ so now we will have to find root of the equation under root.let us assume $\sqrt{8+8\sqrt{3}i}=a+bi$ squaring and comparing real and imaginary part $a^2-b^2=8,2ab=8\sqrt{3}$ so now solving for $a,b$ we get $b=\pm2$ and hence $a=\pm2\sqrt{3}$ so equation becomes $z=\frac{2\sqrt{2}\pm{2\sqrt{3}+2i}}{2i}$ remove out $2$ and multiply and divide by $i$ you get $z=\frac{\sqrt{2}i\pm(\sqrt{3}i-2)}{-1}$ and that are the roots please check my algebra if mistaken but hope you get the method.