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.

## Comments

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TopNewestHaah! 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. – Pi Han Goh · 6 months, 3 weeks ago

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