Author Topic: Complex Number Help!!! (Read 754 times) Tweet Share

batepole · « **on:** October 21, 2013, 04:03:31 pm »

How do i solve z^4 = 2i?!

Alwin · « **Reply #1 on:** October 21, 2013, 04:14:14 pm »

Quote from: batepole on October 21, 2013, 04:03:31 pm

How do i solve z^4 = 2i?!

Hint: put 2i in polar form and then use de moivre's theorem

Oh it's your first post, WELCOME HAHA

EDIT: In the spoiler is the full working just in case the hint wasn't clear enough:

Full working =)

$\begin{aligned} 2i &= 2 \mathrm{cis}\left(\frac{\pi}{2}\right) \\ z^4 &= 2 \mathrm{cis}\left(\frac{\pi}{2} + 2\pi k\right) \text{ where } k\in \mathbb{N} \text{ for 4 solutions} \\ \text{Applying De Moivre's Theorem,} & \\ z&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\frac{\pi}{2} + 2\pi k}{4}\right) \\ z&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8} + \frac{\pi k}{2}\right) \\ k=0, z_1&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}\right) \\ k=1, z_2&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{5\pi}{8}\right) \\ k=2, z_3&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{9\pi}{8}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(-\frac{7\pi}{8}\right) \\ k=3, z_4&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{13\pi}{8}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(-\frac{3\pi}{8}\right) \\ \text{Remember to make sure that } \mathrm{Arg}(z) &\in (-\pi, \pi] \end{aligned}$

Alternate Method =)

$\text{An alternate method is as follows:}$
$\begin{aligned} z^4 &= 2 \mathrm{cis}\left(\frac{\pi}{2} \right) \\ \text{Applying De Moivre's Theorem,} & \\ z&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}\right) \\ \text{We know that there are 4 solutions spaced equally around in a circle} & \\ z_1&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}\right) \\ z_2&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}+\frac{\pi}{2}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(\frac{5\pi}{8}\right) \\ z_3&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}+\pi\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(\frac{9\pi}{8}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(-\frac{7\pi}{8}\right) \\ z_4&=2^{\frac{1}{4}} \mathrm{cis}\left(\frac{\pi}{8}+\frac{3\pi}{2}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(\frac{13\pi}{8}\right) = 2^{\frac{1}{4}} \mathrm{cis}\left(-\frac{3\pi}{8}\right) \\ \text{Remember to make sure that } \mathrm{Arg}(z) &\in (-\pi, \pi] \end{aligned}$

Stevensmay · « **Reply #2 on:** October 21, 2013, 04:21:59 pm »

BasicAcid,
Doesn't $2i = 2 cis(\frac{\pi}{2})$ not $2 cis(\frac{\pi}{4})$

Spoiler

Let $z^4 = 2i = 2 cis(\frac{\pi}{2})$
Applying De Moivre's theorem of $z^n = r^n cis(n \theta )$
$z = \sqrt[4]{2} cis(\frac{\pi}{8})$
As the power of z is 4, we will have four evenly spaced solutions on the complex plane. Thus we add $\frac{2\pi}{4}$ or $\frac{\pi}{2}$ onto our argument to find each solution.
$z_1 = \sqrt[4]{2} cis(\frac{\pi}{8})$
$z_2 = \sqrt[4]{2} cis(\frac{\pi}{8}+ \frac{\pi}{2})$
$z_3 = \sqrt[4]{2} cis(\frac{\pi}{8}+ \frac{2\pi}{2})$
$z_4 = \sqrt[4]{2} cis(\frac{\pi}{8}+ \frac{3\pi}{2})$

Clean up our solutions.
$z_1 = \sqrt[4]{2} cis(\frac{\pi}{8})$
$z_2 = \sqrt[4]{2} cis(\frac{5\pi}{8})$
$z_3 = \sqrt[4]{2} cis(\frac{9\pi}{8})$
$z_4 = \sqrt[4]{2} cis(\frac{13\pi}{8})$

And we are finished.
These cannot be easily converted into Cartesian form, unless there is another component to this question, or calculators are allowed.

Alwin's solution is better. Arguments should generally be within $(\pi,\pi]$ unlike my lazy way of doing it.

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Author Topic: Complex Number Help!!! (Read 754 times) Tweet Share

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