Struggling to solve $\ w^2 + 2iw = i \ $

Question

I'm struggling to solve: $\ w^2 + 2iw = i \ $ I substituted $\ w = x+iy \ $, and eventually got these two equations: $\ x^2 - y^2 = 2y \ $ and $\ 2xy = 1-2x \ $. I'm not quite sure where to go from here. I've tried using $\ x^2 = y^2 + 2y \ $, but I just end up in a dead situation. Can somebody help, please? Thanks

Answer 1

from the equation $$2xy+2x-1=0$$ we obtain $$y=\frac{1-2x}{2x}$$ plug this in your first equation doing this we get $$4x^4+4x^2-1=0$$

Answer 2

Your equation is a quadratic equation. Apply the quadratic formula.

Answer 3

Write as:

$$\begin{align} w^2 +2iw-1 &= i-1 = e^{i 3\pi/4 } \\ (w+i)^2 &= e^{i 3\pi/4 } \end{align}$$

So we get solution as: $-i \pm e^{i3\pi / 8}$ which may be simplified.

Answer 4

Note that we have: $$F(w): w^2 +2iw-i=0$$ Note that we can solve this quadratic in $w$ using the quadratic formula, just as we do it using real numbers.

We thus get: $$w = \frac{ - 2i \pm \sqrt{-4+4i}}{2}$$ $$= - i \pm \sqrt{i-1}$$ as our solutions.