Complex numbers problem

Complex numbers problem

Postby Guest » Sun Apr 01, 2018 12:40 pm

The question is to solve [tex]\sqrt[4]{2\sqrt{3}+2i}[/tex] and the answer is in the attachment, i don't understand how to get that answer.

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Guest
 

Re: Complex numbers problem

Postby Guest » Mon Apr 02, 2018 9:51 pm

Do you know Euler's formula? [tex]e^{i\theta} = \cos \theta + i \sin \theta[/tex]

In a sense, there's two ways to think about complex numbers. There's rectangular where a complex number is represented by its real and imaginary components like [tex]z = a + bi[/tex]. In rectangular, addition is easy, multiplication a little less easy, and exponentiation almost impossible. Then there's polar where it's represented by magnitude and angle like [tex]z = re^{i\theta}[/tex]. Polar representation makes multiplication easy, exponentiation almost as easy, and addition almost impossible. Euler's formula tells you how to convert between rectangular and polar representations.

The plan for this problem is to convert [tex]2\sqrt{3}+2i[/tex] from rectangular to polar, do the 4th root in polar where it's easy, then convert back to rectangular.

The magnitude is given by the Pythagorean identity [tex]r = \sqrt{(2\sqrt{3})^2 + 2^2} = 4[/tex]. The angle is the arctangent of [tex]\frac{2}{2\sqrt{3}} =
\pi/6[/tex].

So [tex]2\sqrt{3}+2i = 4 e^{i\pi/6}[/tex]

Now the 4th root of that is easy, it's just [tex]\sqrt[4]{4} e^{i\pi/24}[/tex]. Actually, there's a small subtlety here because when dividing an angle isn't uniquely specified. Certainly [tex](\pi/6)/4 = \pi/24[/tex], but [tex]\pi/6, 2\pi + \pi/6, 4\pi + \pi/6[/tex], etc are all the same angle, but if you divide each of them by 4, you get different angles depending on how many you wrapped around.


About the answer in your OP, well [tex]\sqrt[4]{4} = \sqrt{2}[/tex], and that's the factor out front. The mess in parentheses is just Euler's formula applied to [tex]e^{i\pi/24}[/tex]. And the stuff about k covers the other solutions arising from the other ways to divide the angle. It looks worse than it is though, because once you've found any 4th root of some complex number, you can get the rest of them by multiplying that solution by all the 4th roots of 1, namely [tex]\pm 1[/tex] and [tex]\pm i[/tex].

Guest
 

Re: Complex numbers problem

Postby Guest » Mon May 14, 2018 1:07 am

Thank you very much.

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