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

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.


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:

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

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

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

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

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

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

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

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

$$ \Leftrightarrow x=\pm 2$$

$$ \Leftrightarrow y= \mp 1 $$

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

  • $\begingroup$ 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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