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# 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
1 year, 6 months ago

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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. · 1 year, 6 months ago