In class today, we learned about complex numbers, and the teacher described a simple procedure for finding the square root of $i$ in $z^2=i$. He explained to us to set $z=a+bi$ and square from there, which did work, but it took some work to get through.

Afterwards, he said that there was an easier way, and that was to use De Moivre's formula to find the square roots.


Question: How would you use De Moivre's formula to find the square root of any nonzero complex number?

I do understand that the formula goes something like this:$$(\cos x+i\sin x)^y=\cos(xy)+i\sin(xy)\tag{1}$$ But I'm not too sure how to apply that to$$z^2=i\tag2$$ for $z$ is another complex number.

  • $\begingroup$ Can you express i in polar form - that is, in terms of cosine and sine? $\endgroup$ – Jack Oct 24 '16 at 3:55
  • $\begingroup$ What it basically says is that raising a complex number to the power n, multiplies the angle from 0 by n. As for the modulus, that is different. $\endgroup$ – The Great Duck Oct 24 '16 at 3:56

Using the $n$-th roots formula (which is actually an equivalent version of De Moivre's formula ) $$ \sqrt{i}=i^{\frac{1}{2}}=(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})^{\frac{1}{2}} =\cos(\frac{\pi}{4}+\kappa\pi)+i\sin(\frac{\pi}{4}+\kappa\pi) $$ for any $\kappa\in\mathbb{Z}$. The above, for $\kappa$ even, gives $$ \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} $$ while for $\kappa$ odd, it gives $$ -\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} $$

For the general case, the formula for the $n$-th roots of the complex number $z=\rho (\cos \phi + i \sin \phi)$, is given by $$ [\rho (\cos \phi + i \sin \phi)]^{1/n} = \rho^{1/n}\left( \cos \frac{\phi + 2 \pi k}{n} + i \sin \frac{\phi + 2 \pi k}{n} \right), \quad k = 0, 1, \dots, $$ $\rho$ is the modulus. For the square roots, just set $n=2$.

  • $\begingroup$ Okay, then how would you use it to find $z^2=2+4i$ for $z$ is a complex number? More generally, how would you use the $n$-th roots formula to find the square root of a complex number where its modulus is not equal to $1$. Could you please add that? $\endgroup$ – Frank Oct 28 '16 at 21:38
  • $\begingroup$ @Frank: I just did. $\endgroup$ – KonKan Oct 29 '16 at 0:06

$i$ has argument $\frac{\pi}{2}$ and modulus $1$, so by de Moivre's formula the number with argument $\frac{\pi}{4}$ and modulus $1$ is a square root of $i$.

Therefore the number $\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})=\frac{1}{\sqrt2}+\frac{i}{\sqrt 2 }$ is a square root for $i$.

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    $\begingroup$ Think you are missing an $i$ in your answer. $\endgroup$ – Oiler Oct 24 '16 at 3:54
  • $\begingroup$ also $\frac{\pi}{4}$ should be in the last line instead of $\frac{\pi}{2}$ $\endgroup$ – KonKan Oct 24 '16 at 3:55
  • $\begingroup$ right, thanks.${}{}$ $\endgroup$ – Jorge Fernández Hidalgo Oct 24 '16 at 3:59
  • $\begingroup$ @JorgeFernándezHidalgo Argument $\frac {\pi}2$ and modulus $1$?? What does that mean? $\endgroup$ – Frank Oct 24 '16 at 4:01
  • $\begingroup$ the argument is the angle formed by the line between the origin and the number ( in the complex plane) and the modules is the distance between the number and the origin in the complex plane. How were you taucgh De-Moivres formula? $\endgroup$ – Jorge Fernández Hidalgo Oct 24 '16 at 4:02

$i$ = $e^{i\frac{\pi}{2}}$, so $\sqrt i $ = $\sqrt {e^{i\frac{\pi}{2}}}$ = $e^{i\frac{\pi}{4}}$


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