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Calculate, in the form $a+ib$, where $a,b\in \Bbb R$, the square roots of $16-30i$.

My attempt with $(a+ib)^2 =16-30i$ makes me get $a^2+b^2=16$ and $2ab=−30$. Is this correct?

Mathxx
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2 Answers2

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I'll give steps. You want $(a+bi)^2 = 16-30i$. Expanding and comparing, you get: $$\begin{cases} a^2-b^2 = 16 \\ 2ab = -30\end{cases}$$

Solve for $b$ in the second equation. Substitute in the first one. You have a second degree equation on $a^2$. Use the quadratic formula to solve for $a^2$. Throw away one of the answers bearing in mind that $a,b \in \Bbb R$. Find two values for $a$. For each one you have a corresponding value of $b$. End.

Ivo Terek
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  • Why I got $a^2+b^2=16$ instead of $a^2-b^2=16$ – Mathxx May 23 '15 at 22:28
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    $$(a+bi)^2 = a^2+2abi+(bi)^2 = a^2+2abi+b^2i^2 = a^2+2abi-b^2$$ – Ivo Terek May 23 '15 at 22:29
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    You can also multiply the first equation by $4a^2$ to obtain $4a^4-(2ab)^2=64a^2$ - and then you have a direct substitution. This is a technique I learned on this site, which seems to lead to simpler manipulations than other approaches. Here it introduces a factor $4$ which can be cancelled, but is otherwise unproblematic. Of course the factor $2$ could be cancelled from the second equation - then just multiply the first by $a^2$. – Mark Bennet May 23 '15 at 22:57
  • @Mark This is nice. I'll start doing it this way from now on :) – Ivo Terek May 23 '15 at 22:59
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    "first equation" $: \mapsto :$ "second equation" $;;;$ ? $;;;;;;;;$ –  May 24 '15 at 01:20
  • Yes sir! My bad. Fixed. – Ivo Terek May 24 '15 at 01:20
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Convert to polar coordinates and write $16-i30=\sqrt{16^2+30^2}e^{-i\arctan(30/16)+2n\pi}$. Upon taking a square root, we find that

$$(16-i30)^{1/2}=(16^2+30^2)^{1/4}e^{-i\frac12 \arctan(30/16)+n\pi}$$

Then, to convert back to Cartesian form, we have

$$(16-i30)^{1/2}=(16^2+30^2)^{1/4}\left(\cos(\frac12 \arctan(30/16)+n\pi)+i\sin(\frac12 \arctan(30/16)+n\pi)\right)$$

Mark Viola
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  • Can you provide more detail for this? How to get $e^{-i\arctan(30/16)+2n\pi}$ ? – Mathxx May 23 '15 at 22:34
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    Sure. No problem. Euler's identity is $e^{iz}=\cos z+i \sin z$. So, upon converting to polar coordinates $x=\rho \cos \theta$ and $y=\rho \sin \theta$. This implies that $z=x+iy=\rho \cos \theta + i\rho \sin \theta=\rho e^{i\theta}$, where $\rho = \sqrt{x^2+y^2}$ and $\theta = \arctan(y/x)$. – Mark Viola May 23 '15 at 22:38
  • Only two values for $n$ suffice, the rest are just repeats. Also, if you want you can further simplify using known identities for $\sin(\arctan(\cdots$, etc. – GPerez May 24 '15 at 00:26