Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

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fp2 complex equation

solve z^5 = 4 + 4i, giving answers in the form z=e^( ik(pi) ) where k is rational and 0<=k<=2

why are some answers of theta out of the range -pi<theta<=pi ?
i dont get it

Original post by cooldudeman

solve z^5 = 4 + 4i, giving answers in the form z=e^( ik(pi) ) where k is rational and 0<=k<=2

why are some answers of theta out of the range -pi<theta<=pi ?
i dont get it

If you have a complex number

r e^{i \theta + 2 \pi k}

and you want to find the nth roots of unity, your nth roots of unity will be

r^{1/n} e^{\frac{i \theta + 2 \pi k}{n}}

where

0 \leq k \leq (n-1)

Your argument for the complex number doesn't necessarily have to be expressed between -pi and pi

OP

Original post by Felix Felicis

If you have a complex number

r e^{i \theta + 2 \pi k}

and you want to find the nth roots of unity, your nth roots of unity will be

r^{1/n} e^{\frac{i \theta + 2 \pi k}{n}}

where

0 \leq k \leq (n-1)

Your argument for the complex number doesn't necessarily have to be expressed between -pi and pi

so you just keep adding 2pi until you have 'n' amount of answers?

and how do you know whether or not you have negative answers of theta?

(edited 11 years ago)

Original post by cooldudeman

so you just keep adding 2pi until you have 'n' amount of answers?

Yes, but you go up to (n-1) as you start with k=0.

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