Wolfram Alpha doesn't show all solutions for square root of complex number?

Question

I found the square root of $3-4i$, getting the results $-2+i$ and $2-i$. But when I put this into Wolfram|Alpha, it only showed the solution $2-i$. Is this an error on my part, on Wolfram Alpha's part, or am I just missing something?

Answer 1

Because $3$ and $4$ are each divisible by only one small prime factor, by comparing coefficients (N.B. $i^2=-1$), $$3-4i=(2-i)(2-i)$$ and $$3-4i=(i-2)(i-2)$$.

The other answer shows the standard general method by equating real and imaginary parts, but it is way too long for this question.

Answer 2

What is the stuff with the tangents and $e$

Using polar form: $$ z^2 = 3-4i $$

$$ \Leftrightarrow z^2 = 5e^{i \cdot (\tan^{-1}(\frac{-4}{3}))} $$

$$ \Leftrightarrow z^2 = 5e^{i \cdot (\tan^{-1}(\frac{-4}{3})+2\pi k)} $$

$$ \Leftrightarrow z= \sqrt{5}e^{i \cdot (\frac{\tan^{-1}(\frac{-4}{3})}{2}+\pi k)} $$

$$ \therefore z_1 = \sqrt{5}e^{i \cdot (\frac{\tan^{-1}(\frac{-4}{3})}{2}+\pi )} , z_2=\sqrt{5}e^{i \cdot (\frac{\tan^{-1}(\frac{-4}{3})}{2})} $$

Alternative way:

$$ z^2 = 3-4i $$

$$\Leftrightarrow (x+iy)^2 = 3-4i $$

$$ \Leftrightarrow (x^2-y^2) +(2xy)i = 3-4i$$

$$ \Leftrightarrow x^2-y^2=3 ~~~~, ~~~ 2xy=-4 ~\left(y= \frac{-2}{x}\right) $$

$$ \Leftrightarrow x^2-\left(\frac{-2}{x}\right)^2=3 $$

$$ \Leftrightarrow x^2 - \frac{4}{x^2} = 3$$

$$ \Leftrightarrow x^4-3x^2-4=0 $$

$$ \Leftrightarrow x^2 = 4 ~( x \in R ) $$

$$ \Leftrightarrow x=\pm 2$$

$$ \Leftrightarrow y= \mp 1 $$

$$ \therefore z_1 = -2+i , z_2 = 2-i $$