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 Exercise [05.14] 
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Joined: 22 May 2008, 19:08
Posts: 18
Post Exercise [05.14]
z=w^{1/n}=\left(re^{i\theta+2k\pi i}\right)^{1/n}=\sqrt[n]{r}\cdot e^{\frac{i\theta+2k\pi i}{n}}=\sqrt[n]{r}\cdot e^{\frac{i\theta}{n}}\cdot e^{\frac{2k\pi i}{n}}=z_0\cdot e^{\frac{2k\pi i}{n}}

All n solutions to the equation will be given by z_0\cdot e^{\frac{2k\pi i}{n}} for k=0,...,n-1.

P.S.: Missing \frac1n corrected. Thanks to vasco for noticing.


Last edited by ZZR Puig on 08 Jul 2008, 09:06, edited 2 times in total.

08 Jun 2008, 18:49
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [05.14]
Hi ZZR
Just a small correction to your solution, which I'm sure was just a typing error. The penultimate expression is missing a division by n in the exponent. So it should be:

z=w^{1/n}=\left(re^{i\theta+2k\pi i}\right)^{1/n}=\sqrt[n]{r}\cdot e^{\frac{i\theta+2k\pi i}{n}}=\sqrt[n]{r}\cdot e^{\frac{i\theta}{n}}\cdot e^{\frac{2k\pi i}{n}}=z_0\cdot e^{\frac{2k\pi i}{n}}

All n solutions to the equation will be given by z_0\cdot e^{\frac{2k\pi i}{n}} for k=0,...,n-1.


08 Jul 2008, 07:17

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [05.14]
One way this could be written more in keeping with the "when successively different values of the logw are specified" is perhaps:

Image
Image



and so on similarly for n=3, this function is 3-fold period and 3 values are bounced around ad infinitum.

Hence we only need assume k=0,....,n-1 to arrive at all the distinct values.


22 Sep 2011, 10:27
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