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This question already has an answer here:

How do I find the square root of complex number $7-(6\sqrt2)i$?

I hope there's someone who can show me the method. Thanks in advance.

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marked as duplicate by David K, Hans Lundmark, iadvd, Shailesh, Eric Stucky Oct 5 '16 at 2:12

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  • $\begingroup$ There is no "THE squareroot of a complex number". It is better to write "Solve the equation $x^2=7-(6\sqrt{2})i$ for $x$ and you get two answers, neither of them being positive or negative since they also will be...complex $\endgroup$ – imranfat Oct 4 '16 at 17:40
  • $\begingroup$ We'll need to write this in the form $re^{i\theta}$ and use de Moivre. $\endgroup$ – Sean Roberson Oct 4 '16 at 17:40
  • $\begingroup$ @TahaAkbari Good catch $\endgroup$ – imranfat Oct 4 '16 at 17:41
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    $\begingroup$ try $a + b i \sqrt 2,$ squre it and see if you get $a,b$ integers or at least rational. $3 - i \sqrt 2$ works, did it in my head $\endgroup$ – Will Jagy Oct 4 '16 at 17:42
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    $\begingroup$ @WillJagy I worked it out, got the answers and indeed the imaginary component has a square root. What is going on in your head:) ? $\endgroup$ – imranfat Oct 4 '16 at 17:45
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$$\sqrt{7-6\sqrt{2}i}=\sqrt{7+2-2-6\sqrt{2}i}=\sqrt{9-6\sqrt{2}i-2}=\sqrt{3^2-2*3\sqrt{2}i+(\sqrt2i)^2}=\sqrt{(3-\sqrt{2}i)^2}$$

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  • $\begingroup$ Nice way, but honestly, you got to see that one somehow...There are also two solutions from your square root. $\endgroup$ – imranfat Oct 4 '16 at 17:51

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