- #1

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Jut had a question. How do I go about finding the general formula for roots of the complex poly [tex]{z}^{n}-a[/tex] where a is another complex number.

Do I just go [tex]{z}^{n}=a[/tex]??? :S so complicated this things!!!

Thanks in advance!

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- Thread starter complexhuman
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- #1

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Jut had a question. How do I go about finding the general formula for roots of the complex poly [tex]{z}^{n}-a[/tex] where a is another complex number.

Do I just go [tex]{z}^{n}=a[/tex]??? :S so complicated this things!!!

Thanks in advance!

- #2

lurflurf

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It is easier if z is written in geometric formcomplexhuman said:

Jut had a question. How do I go about finding the general formula for roots of the complex poly [tex]{z}^{n}-a[/tex] where a is another complex number.

Do I just go [tex]{z}^{n}=a[/tex]??? :S so complicated this things!!!

Thanks in advance!

let a=u+i v where u and v are real

z=r(cos(t)+i sin(t)) where 0<=r

then

z^n-a becomes

(r^n)(cos(n t)+i sin(n t))-a

so r should be chosen by

r^n=|a|=sqrt(a^2+b^2)

cos(n t)+i sin(n t)=a/|a|

so

cos(n t)=u/|a|

sin(n t)=v/|a|

tan(n t)=v/u

t=(1/n)Arctan(v/u)

this is one solution but n can be found using the usual methods of solving trigonometric equations.

in fact

t=(1/n)Arctan(v/u)+2pi k/n

k=0,1,2,...,n-2,n-1,

thus

z=r(cos(t)+i sin(t))

=|a|^(1/n)(cos((1/n)Arctan(v/u)+2pi k/n)+i sin((1/n)Arctan(v/u)+2pi k/n

))

k=0,1,2,...,n-2,n-1

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- #3

AKG

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It's actually easiest to understand in polar co-ordinates. [itex]a = |a|\exp (i\theta )[/itex]. So simply take |z| to be |a|^{1/n}. Take angles [itex]\theta _k = \frac{\theta + 2k\pi}{n}[/itex] for values k = 0, 1, ..., n-1. Then your roots are the complex numbers [itex]|z|\exp (i\theta _k)[/itex] for the n values for k.

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HallsofIvy

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- #5

AKG

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- #6

HallsofIvy

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Writing a+ bi as r(cos(θ)+ i sin(θ)) **is** writing the complex number in polar form.

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AKG

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HallsofIvy

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