I am asked the following question:

Find the square roots of $z = 5-12i$

I know that this problem can be easily solved by doing the following:

$$ z_k^2 = 5-12i\\ (a+bi)^2 = 5-12i\\ (a^2-b^2) + i(2ab) = 5-12i\\ \\ \begin{cases} a^2 - b^2 = 5\\ 2ab = -12 \end{cases} \quad \Rightarrow \quad z_1 = -3+2i \quad z_2 = 3-2i $$

My question is: can the following method (below) be used to solve the problem above? Motivation for this question: if I were to find the cubic roots of the number given, I couldn't use the first method.

I will use this "other method" it in a different problem.

Find the square roots of $ z = 2i $

The method:

Since $ \rho = 2 $ and $ \theta = \frac{\pi}{2} $ we have to find a complex number such that

\begin{align*} z_k^2 &= 2i\\ z_k^2 &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\ \rho_k^2 \left( \cos 2\theta_k + i \sin 2 \theta_k \right) &= 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)\\ \end{align*}

\begin{cases} \rho^2 &= 2\\ 2\theta_k &= \frac{\pi}{2} + 2k\pi \end{cases}

\begin{cases} \rho &= \sqrt{2}\\ \theta_k &= \frac{\pi}{4} + 2k\pi \end{cases}

\begin{align*} z_0 &= \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 1+i\\ z_1 &= \sqrt{2} \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = -1-i \end{align*}

Using this method for the first example seems like a dead end, specially because of the fact that the angle of $z$ is not as straightforward as the angle of the second example.

Thank you.

  • $\begingroup$ Your second method appears to be using DeMoivre's theorem, It's very effective in finding powers and roots of complex numbers, and is quicker at roots than your first method. $\endgroup$ – tberkeley Sep 20 '16 at 20:36
  • $\begingroup$ Yes it is very useful, but I am wondering how useful it is on cases when the angles are not on the 30-45-60-90-180-... range. Can you answer that tberkeley? $\endgroup$ – bru1987 Sep 20 '16 at 20:37
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    $\begingroup$ It's a good question. I'm sorry I can't answer it. $\endgroup$ – tberkeley Sep 20 '16 at 20:51
  • $\begingroup$ If $z = a + bi $ then find $\theta = \arctan b/a$ and let $r = \sqrt {a^2+b^2}$. Then $z =r (\cos \theta + \sin \theta)$ and $z {1/n}= \sqrt [n]{r}(\cos( \theta/n+2k\pi/n) +i\sin (\theta/n + 2k\pi/n)) $ $\endgroup$ – fleablood Sep 20 '16 at 21:51

If you let $\theta$ be the angle in the first case, then using the fact that $\tan \theta = -12/5$, you can find $\cos\theta$ and $\cos\frac{\theta}{2}$ indirectly. One of the roots would be: $\sqrt{13}(-\frac{3}{\sqrt{13}} + i \frac{2}{\sqrt{13}}) = z_1.$

  • $\begingroup$ you mean finding cos and sin using for example $\cos^2x+\sin^2x = 1$? $\endgroup$ – bru1987 Sep 20 '16 at 20:53
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    $\begingroup$ @bru1987 Yes! $1+\tan^2x = \frac{\cos^2x + \sin^2x}{\cos^2x} = \frac{1}{\cos^2x}$, and so $\frac{1}{\cos^2x} = 1+(12/5)^2$. From here, you can solve for $\cos x$ and $\sin x$. $\endgroup$ – trang1618 Sep 20 '16 at 21:10

All you need to do is to convert your $z$ to polar form:

$$z=5-12i=\sqrt{12^2+5^2}e^{i(\arctan(-\frac{12}{5})+k\pi)}=13e^{i(-1.176+k\pi)}$$ Hence, one answer of $z^{\frac{1}{2}}$ is $$z_1=13^{\frac{1}{2}}e^{i\frac{-1.176}{2}}=\sqrt{13}e^{i(-0.588)}$$

Converting this to Cartesian will give: $$z_1=\sqrt{13}\left(\cos(-0.588)+i\sin(-0.588)\right)=3-2i$$ which is one of your original results.


You can apply de Moivre's theorem (second example) on the first example without doubt. Though you need a calculator or a trigonometrical table handy to check the sin and cos of a certain angle which feels kinda painful, at least for me.


Your method should work fine in both cases (though the trigonometry is a bit more complicated), but I point out that your first method can still work on the square roots of $2i$: We see that

$$ a^2-b^2 = 0 $$

so $a = \pm b$, and then, secondly, $2ab = 2$. This gives us $a = b = \pm 1$ as the solutions, so the square roots of $2i$ are $1+i$ and $-1-i$.


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