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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?

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    $\begingroup$ Wolfram alpha is just listing the principal square root. If you scroll down, it says "all 2nd roots of $3-4i$ and lists both... same thing happens if you ask Wolfram alpha for the square root of, say, $9$. $\endgroup$ – kccu Nov 4 '16 at 4:25
  • $\begingroup$ Oh I see, thanks. What is the stuff with the tangents and $e$? I didn't use any of that to solve this $\endgroup$ – suomynonA Nov 4 '16 at 4:27
  • $\begingroup$ @suomynonA How did you solve it? $\endgroup$ – bigfocalchord Nov 4 '16 at 4:28
  • $\begingroup$ If $z=re^{i \theta}$ with $-\pi < \theta \leq \pi$ (this is the polar form of the complex number), then the principal square root of $z$ is $\sqrt{r}e^{i \theta/2}$. Wolfram alpha seems to have done some simplification to get $\arctan(\frac{1}{2})$, since $\frac{1}{2} \neq \tan \theta$ when we write $3-4i$ in polar form... $\endgroup$ – kccu Nov 4 '16 at 4:40
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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.

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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:

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

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

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

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

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

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

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

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

$$ \Leftrightarrow x=\pm 2$$

$$ \Leftrightarrow y= \mp 1 $$

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

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  • $\begingroup$ I used the "alternative" way; that's why I was confused. $\endgroup$ – suomynonA Nov 4 '16 at 5:24
  • $\begingroup$ @suomynonA Yeah the first method also gives the same answer if you convert it back to rectangular form. $\endgroup$ – bigfocalchord Nov 4 '16 at 5:28

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