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: zz be a complex number in the form a+bi\boxed{a + bi}, nn be an integer where n≠0\boxed{n \neq 0}

zn\large \sqrt[n]{z}

z1n\large z^{\frac{1}{n}}

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

r1neθin\large r^{\frac{1}{n}}e^{\frac{\theta i}{n}}

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

Note: xx is an integer between 00 and n−1n - 1 inclusive. The reason it's in there is to find all of the roots of zz also known as the roots of unity (no not the game engine). Anyways onwards with the calculation.

a2+b22n cis (arctan⁡ba+2xπn)\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 zn=Zx\large \sqrt[n]{z} = Z_x

Zx=a2+b22n cis (arctan⁡ba+2xπn)\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.

#Algebra

Note by Jack Rawlin
5 years, 3 months ago

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Comments

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.

Pi Han Goh - 5 years, 3 months ago
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