Fourth root of a complex number. (De Moivre's Formula)

Algebra

Fourth root of a complex number. (De Moivre's Formula)

Postby alibabakes5 » Fri Sep 17, 2021 9:24 am

I tried looking for help to understand this topic, but I still can't even grasp the concept.

I understood the formula, but not the roots. The teacher says that each complex number has 4 fourth roots (as well as 3 third roots, 5 fifth roots, etc.) and things like that, but I never knew how to calculate that, and I need to know since we have an exam on friday. Also, I can't really ask the teacher, since his classes are recorded (we don't ever see him in person).
alibabakes5
 
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Re: Fourth root of a complex number. (De Moivre's Formula)

Postby Baltuilhe » Tue Sep 21, 2021 10:00 pm

Sample:

[tex]z=a+bi[/tex]
or
[tex]z=\rho\cdot\left(\cos\theta+i\sin\theta\right)[/tex]

So:
[tex]a=\rho\cos\theta\\
b=\rho\sin\theta[/tex]
and
[tex]\tan\theta=\frac{b}{a}\\\\
\rho=\sqrt{a^2+b^2}[/tex]

De Moivre's Root Theorem:
[tex]z^{1/n}=\rho^{1/n}\cdot\left[\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right)\right][/tex]
k=0,1,2,...,n-1

Numeric:
[tex]z=16\cdot\left[\cos\left(\dfrac{\pi}{6}\right)+i\sin\left(\dfrac{\pi}{6}\right)\right][/tex]

Fourth roots:
[tex]z^{1/4}=16^{1/4}\cdot\left[\cos\left(\dfrac{\dfrac{\pi}{6}+2k\pi}{4}\right)+i\sin\left(\dfrac{\dfrac{\pi}{6}+2k\pi}{4}\right)\right][/tex]

[tex]z^{1/4}=(2^4)^{1/4}\cdot\left[\cos\left(\dfrac{\pi+12k\pi}{24}\right)+i\sin\left(\dfrac{\pi+12k\pi}{24}\right)\right][/tex]
k=0
[tex]z^{1/4}=2\cdot\left[\cos\left(\dfrac{\pi}{24}\right)+i\sin\left(\dfrac{\pi}{24}\right)\right][/tex]
k=1
[tex]z^{1/4}=2\cdot\left[\cos\left(\dfrac{13\pi}{24}\right)+i\sin\left(\dfrac{13\pi}{24}\right)\right][/tex]
k=2
[tex]z^{1/4}=2\cdot\left[\cos\left(\dfrac{25\pi}{24}\right)+i\sin\left(\dfrac{25\pi}{24}\right)\right][/tex]
k=3
[tex]z^{1/4}=2\cdot\left[\cos\left(\dfrac{37\pi}{24}\right)+i\sin\left(\dfrac{37\pi}{24}\right)\right][/tex]

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