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Exercise [05.14]
http://www.roadtoreality.info/viewtopic.php?f=19&t=164
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Author:  ZZR Puig [ 08 Jun 2008, 18:49 ]
Post subject:  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.

Author:  vasco [ 08 Jul 2008, 07:17 ]
Post subject:  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.

Author:  Azrael84 [ 22 Sep 2011, 10:27 ]
Post subject:  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.

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