MHB What Are the Solutions to the Equation \(x^2 + 2i = 0\)?

[karush]

$${x}^{2}+2i=0$$
$$\left(x-? \right)\left(x-? \right)=0$$
This should be easy but I couldn't get the factor

 

[Fernando Revilla]

First way. Using trigonometric form :
$$x^2=-2i\Leftrightarrow x=\sqrt{-2i}=\sqrt{2[\cos 3\pi/2+i\sin 3\pi/2]}=\ldots= \pm (1-i).$$
Second way. Using the definition of square root. If $x=x_1+x_2i$ with $x_1,x_2\in\mathbb{R},$ then
$$x^2=-2i\Leftrightarrow (x_1+x_2i)^2=-2i\Leftrightarrow x_1^2+2x_1x_2 i-x_2^2=-2i$$ $$\Leftrightarrow \left \{ \begin{matrix} \displaystyle\begin{aligned} & x_1^2-x_2^2=0\\& 2x_1x_2=-2 \end{aligned}\end{matrix}\right.\Leftrightarrow \ldots\Leftrightarrow \left \{ \begin{matrix} \displaystyle\begin{aligned} & x_1=1,\;x_2=-1\\& x_1=-1,\;x_2=1 \end{aligned}\end{matrix}\right.\Leftrightarrow x=\pm(1-i).$$
Third way. By simple inspection $(1-i)^2=1-2i-1=-2i$ so, also $[-(1-i)]^2=-2i.$

That is, $x^2+2i=[x-(1-i)][x-(-1+i)].$

 

[karush]

I like the 3rd way

But looks like next problem is trig form

 

[Fernando Revilla]

I like the 3rd way

So do I. :)

But looks like next problem is trig form

In that sense, I can't help you. :)