Polar Form Question

2+2√3i in polar form?
I know about how to do it if it were something as simple as 3+3 but when it is made a little more complicated it gets tricky for me.

I don't want any answers telling me to google it, believe me I have looked...I even bought a precalc for dummies book which has been helpful but it doesn't explain this very well. I just need a break down step by step so I know how to solve like problems.

Polar form
z = 2 + 2√3i = r(cosΘ + isinΘ)

r = √[2² + (2√3)²] = √(4+12) = √16 = 4

z = 4(1/2 + 2√3i/4) = 4(cosΘ + isinΘ)

Comparing the real and imaginary parts, we see that:
1/2 = cosΘ, √3/2 = sinΘ
therefore, Θ = arccos(1/2) = arcsin(√3/2) = π/3
and

z = 4[cos(π/3) + i sin(π/3)]
or
z = 4 cis π/3
or
z = 4e^(iπ/3)

Any complex number
"x+yi" can be represented in the polar form as
r(cosΘ+isinΘ)
or rcisΘ
or r<Θ (Note: Here "<" denotes the symbol used for angles)
or re^(iΘ)
where r=sqrt(x^2+y^2) and Θ=arctan(y/x)

For 2+2sqrt(3)i,
r=sqrt{2^2+(2sqrt(3))^2} i.e, r=sqrt(4+12)=4

Θ=arctan{2sqrt(3)/2}
i.e, Θ=arctan{sqrt(3)} =Π/3.

(Note: Here Θ lies in 1st quadrant since both "x" and "y" for the given complex number are positive. Similarly, depending on the signs of x and y in "x+yi", you can ascertain the quadrant in which Θ lies.)

Hence, the required polar form is any of the following:

4{cos(Π/3)+isin(Π/3)}

or 4cis(Π/3)

or 4<Π/3 (here the symbol '<' denotes the symbol we use for angles).

Or 4e^(iΠ/3).