# Complex roots

Recently I was looking back over some of my previous posts to see if I could improve them. While doing this I noticed something; I haven't done complex roots yet. This surprised me since it's the easiest to calculate by far (at least from what I've found).

So let's get on with it.

Let: $z$ be a complex number in the form $\boxed{a + bi}$, $n$ be an integer where $\boxed{n \neq 0}$

$\large \sqrt[n]{z}$

$\large z^{\frac{1}{n}}$

$\large \left(re^{\theta i}\right)^{\frac{1}{n}}$

$\large r^{\frac{1}{n}}e^{\frac{\theta i}{n}}$

$\large \sqrt[n]{r} \text{ cis }\left(\frac{\theta + 2x\pi}{n}\right)$

Note: $x$ is an integer between $0$ and $n - 1$ inclusive. The reason it's in there is to find all of the roots of $z$ also known as the roots of unity (no not the game engine). Anyways onwards with the calculation.

$\large \sqrt[2n]{a^2 + b^2} \text{ cis }\left(\frac{\arctan{\frac{b}{a} + 2x\pi}}{n}\right)$

Like I said, easiest to calculate by far.

Let $\large \sqrt[n]{z} = Z_x$

$\large Z_x = \sqrt[2n]{a^2 + b^2} \text{ cis }\left(\frac{\arctan{\frac{b}{a} + 2x\pi}}{n}\right)$

Hope you enjoyed the note.

Note by Jack Rawlin
3 years, 8 months ago

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## Comments

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Haah! Another nice note! I bookmarked this previously so I can read them on my free time.

You should paste this in one of the De Moivre's Theorem's wiki.

- 3 years, 8 months ago

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