# Trigonometric Complex Numbers

1. Homework Statement
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how are the solutions of the fourth roots pi/2? how do you get pi/2, you know the thing after "cos" and "isin" ?

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First write it in polar form, or trigonometric form, however you call it, and after that use de moivre's formula to find its roots.

let z be a complex nr.

z=a+bi, writing it in polar forms : $$a=r cos(\theta),b=\ro sin\theta$$

So,

$$z=r (cos\theta+isin\theta)$$

now

$$z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos\frac{\theta +2k\pi}{n}+isin{\frac{\theta+2k\pi}{n})$$

Now all you need to do is figure out what $$\theta$$ is, and your fine.

Or if you want the exponential representation of a complex nr:

$$e^{ix}=cosx+isinx$$

$$e^{i\frac{\pi}{2}}=i$$ so we get

$$i=cos\frac{\pi}{2}+isin\frac{\pi}{2}$$

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$$e^{i\frac{\pi}{2}}=i$$
how do you know that x = pi/2

well z=i, is a complex nr right. Following my elaboration above we have a=0, b=1, right?

so $$\theta =\arctan\frac{1}{0}-->\frac{\pi}{2}$$ loosly speaking.

so the exponential form of i is what i wrote $$i=e^{i\frac{\pi}{2}}$$

SO to find the four roots of i just follow de moivres formula that i wrote above, letting k=0,1,2,3.

how do you know a=0 and b=1? isnt it a=1 and b=-1 or did i do something wrong

HallsofIvy