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

[alibabakes5]

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).

[Baltuilhe]

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]

[Guest]

For another example, consider the fourth roots of the real number "16". since cos(0)= 1 and sin(0)= 0 we can write that as 16(cos(0)+ i sin(0)). By "DeMoivres' Theorem", the four fourth roots are 16^(1/4)(cos(0+ 2npi/4+ isin(0+ 2npi/4))= 16^(1/4)(cos(npi/2)+ isin(npi/2)).

The "16^(1/4)" means specifically the real fourth root and, since 2(2)(2)(2)= 4(2)(2)= 8(2)= 16, 16^(1/4)= 2.

When n= 1 cos(pi/2)+ isin(pi/2)= 0+ i so 16^(1/4)(cos(npi/2)+ isin(npi/2))= 2(0+ i)= 2i. You can check that (2i)^4= (2i)(2i)(2i)(2i)= (-4)(-4)= 16.

When n= 2 cos(pi)+ isin(pi)= -1+ 0i= -1 so 16^(1/4)(cos(npi/2)+ isin(npi/2))= 2(-1+ 0i)= -2. You can check that (-2)^4= (-2)(-2)(-2)(-2)= (4)(4)= 16.

When n= 3 cos(3pi/2)+ isin(3pi/2)= 0+ i so 16^(1/4)(cos(npi/2)+ isin(npi/2))= 2(0- i)= -2i. You can check that (-2i)^4= (-2i)(-2i)(-2i)(-2i)= (-4)(-4)= 16.

When n= 4 cos(4pi/2)+ isin(4pi/2)= cos(2pi)+ i sin(2pi)= 1+ 0i=so 16^(1/4)(cos(npi/2)+ isin(npi/2))= 2(1+ 0i)= 2. You can check that (2)^4= (2)(2)(2)(2)= (4)(4)= 16.

If you were to continue on, to n= 5, 6, etc., because sine and cosine are periodic with period 2pi, you would just get the same four roots again.