Find a polar form, z=re^iθ, of the complex number z=-6+2 sqrt 3 i. Use exact values?

Recall:

z = a + ib ← this is a complex number

m = √(a² + b²) ← this is its modulus

tan(α) = b/a → then you can deduce α ← this is the argument

In your case:

z = - 6 + 2i√3 ← you can see that: a = - 6 and you can see that: b = 2√3

m = √[(- 6)² + (2√3)²]

m = √[36 + 12]

m = √48

m = 4√3 ← this is the modulus of z

tan(α) = b/a

tan(α) = (2√3)/- 6

tan(α) = - (√3)/3

α = π - (π/6)

α = 5π/6

z = (4π3).e^[(5π/6).i]

z = r * e^(i * t)

z = r * (cos(t) + i * sin(t))

z = r * cos(t) + r * sin(t) * i

z = -6 + 2 * sqrt(3) * i

r * cos(t) = -6

r * sin(t) = 2 * sqrt(3)

r^2 * cos(t)^2 = 36

r^2 * sin(t)^2 = 4 * 3 = 12

r^2 * cos(t)^2 + r^2 * sin(t)^2 = 36 + 12

r^2 * (cos(t)^2 + sin(t)^2) = 48

r^2 * 1 = 48

r = sqrt(48)

r = sqrt(16 * 3)

r = -4 * sqrt(3) , 4 * sqrt(3)

Let's assume a positive value for r (we'll do the negative value later)

r * cos(t) = -6

4 * sqrt(3) * cos(t) = -6

cos(t) = -6 / (4 * sqrt(3))

cos(t) = -6 * sqrt(3) / (4 * 3)

cos(t) = -6 * sqrt(3) / 12

cos(t) = -sqrt(3) / 2

t = 5pi/6 , 7pi/6

4 * sqrt(3) * sin(t) = 2 * sqrt(3)

sin(t) = 1/2

t = pi/6 , 5pi/6

5pi/6 is the common value of t

4 * sqrt(3) * (cos(5pi/6) + i * sin(5pi/6)) =>

4 * sqrt(3) * e^((5pi/6) * i)

Using a negative value for r

-4 * sqrt(3) * cos(t) = -6

cos(t) = sqrt(3)/2

t = pi/6 , 11pi/6

-4 * sqrt(3) * sin(t) = 2 * sqrt(3)

sin(t) = -1/2

t = 7pi/6 , 11pi/6

t = 11pi/6

z = -4 * sqrt(3) * e^((11pi/6) * i)

Hopefully you can see the symmetry here. By multiplying r by -1, our angle is out by pi radians or 180 degrees.

z = -6 + 2 sqrt 3 i

z = √((-6)² + (2 sqrt 3)²) (-6 + 2 sqrt 3 i) / √((-6)² + (2 sqrt 3)²)

z = 4√3 (-6 + 2 sqrt 3 i) / (4√3)

z = 4√3 (-√3/2 + 1/2 i)

z = 4√3 cis(5π/6)

z = 4√3 e^(i 5π/6)

try mathway or photomath