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What are the fourth roots of -3+3sqrt3i ?

 Apr 10, 2020
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Here we make use of de moivre's theorem, which states that given a complex number z:

 

\(z=re^{i\theta} = r(\cos{\theta} + i\sin{\theta})\)

We can then raise z to the nth power to get a general form for this:

\(z^n = r^ne^{i*n*\theta} = r(\cos{n\theta} + i\sin{n\theta})\)

We can then set z as our desired complex number. We then get:

 

\(-3+3i\sqrt{3} = 6(\cos{2\pi\over3} + i\sin{2\pi\over3})\)

More generally, we can write:

\(-3+3i\sqrt{3} = 6(\cos(2n\pi +{2\pi\over3}) + i\sin(2n\pi + {2\pi\over3}))\)

\((-3+3i\sqrt{3})^{1/4} = 6(\cos(2n\pi/4 +{2\pi\over12}) + i\sin(2n\pi/4 + {2\pi\over12}))\)

Simplifying, we get:

\((-3+3i\sqrt{3})^{1/4} = 6(\cos(n\pi/2 +{\pi\over6}) + i\sin(n\pi/2 + {\pi\over6}))\)

Now we get our 4 roots from testing n from 0-3(it iterates after a cycle of 4). I'm sure you're capable enough to plug and chug in the numbers! :)

 

Edit: here's a good reference; it uses a similar method and explanation

https://socratic.org/questions/how-do-you-find-the-fourth-roots-of-i

 Apr 10, 2020
edited by jfan17  Apr 10, 2020

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