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

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The equation \(x^2 + 2i = 0\) can be solved using various methods, all leading to the same solutions. The trigonometric form yields \(x = \pm(1-i)\) by expressing \(-2i\) in polar coordinates. Another approach involves expanding \(x = x_1 + x_2i\) and equating real and imaginary parts, also resulting in \(x = \pm(1-i)\). A straightforward inspection confirms that \((1-i)^2 = -2i\), validating the solutions. Ultimately, the solutions to the equation are \(x = 1-i\) and \(x = -1+i\).
karush
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$${x}^{2}+2i=0$$
$$\left(x-? \right)\left(x-? \right)=0$$
This should be easy but I couldn't get the factor
 
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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, $x^2+2i=[x-(1-i)][x-(-1+i)].$
 
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I like the 3rd way

But looks like next problem is trig form
 
karush said:
I like the 3rd way
So do I. :)
But looks like next problem is trig form
In that sense, I can't help you. :)
 
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