Originally Posted by

**HallsofIvy** Actually, you did NOT make an error, you simply have not finished the problem. You want to write the solutions in the "standard" form "a+ bi".

Use the quadratic **formula** (not "quadratic equation") $\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ with a= 1, b= 4i, and c= -4- 2i.

$\displaystyle \frac{-4i\pm\sqrt{(-4i)^2- (4)(1)(-4- 2i)}}{2(1)}= \frac{-4i\pm\sqrt{-16+ 16+ 8i}}{2}= \frac{-4i\pm\sqrt{8i}}{2}$

$\displaystyle = -2i+ \sqrt{2i}$ (which is what you got) where I have dropped the "$\displaystyle \pm$" since I will be using the two roots of 2i.

"2i" is at (0, 2) in the complex plane, 90 degrees from the real axis, so its square roots are at 45 and -45 degrees, still distance 2 from the origin. That is, they are $\displaystyle \sqrt{2}(1+ i)$ and $\displaystyle \sqrt{2}(1- i)$. The solutions to the equation are

$\displaystyle -2i+ \sqrt{2}(1+ i)= \sqrt{2}+ (\sqrt{2}- 2)i$ and $\displaystyle -2i+ \sqrt{2}(1- i)= \sqrt{2}- (\sqrt{2}+ 2)i$.