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

Algebra

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

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

Posts: 3
Joined: Mon Aug 03, 2020 5:57 am
Reputation: -50

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

Sample:

$$z=a+bi$$
or
$$z=\rho\cdot\left(\cos\theta+i\sin\theta\right)$$

So:
$$a=\rho\cos\theta\\ b=\rho\sin\theta$$
and
$$\tan\theta=\frac{b}{a}\\\\ \rho=\sqrt{a^2+b^2}$$

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

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

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

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

Baltuilhe

Posts: 98
Joined: Fri Dec 14, 2018 3:55 pm
Reputation: 63