Author Topic: Solving Question (Read 792 times) Tweet Share

sameeehh · « **on:** January 15, 2013, 11:24:48 pm »

Hey guys , i'm just trying to do some holiday homework before school starts and require some help.

Q. The solutions to z^3 = root 3 + i in Polar form are?

I'm kind of just following rules atm, however a little bit of explanation would be greatly appreciated .
cheers guys !

brightsky · « **Reply #1 on:** January 15, 2013, 11:45:44 pm »

express sqrt(3) +i in polar form and then use de moivre's theorem. the steps are basically the same for all equations of form z^n = a.

polar · « **Reply #2 on:** January 16, 2013, 11:03:18 am »

$\begin{aligned} z^3 & = \sqrt{3} + i <br />\\ \text{Express \;} \sqrt{3} + i \; \text{in polar form:}<br />\\ z^3 & = 2\mathrm{cis}\left(\frac{\pi}{6} + 2\pi{n}\right), \; n \in Z<br />\\ \text{Apply de Moivre's Theorem: if \;} z = r\mathrm{cis}(\theta), z^n = r^n\mathrm{cis}(\theta)<br />\\ z & = 2^{\frac{1}{3}}\mathrm{cis}\left(\frac{\pi}{18} + \frac{2\pi{n}}{3}\right) \end{aligned}$

hence, the solutions are $z = 2^{\frac{1}{3}}\mathrm{cis}\left(\frac{\pi}{18}\right), 2^{\frac{1}{3}}\mathrm{cis}\left(\frac{13\pi}{18}\right), 2^{\frac{1}{3}}\mathrm{cis}\left(\frac{25\pi}{18}\right)$

sameeehh · « **Reply #3 on:** January 17, 2013, 07:54:44 pm »

thanks !! appreciate it !

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