la console
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]
Captain Matticus, LandPiratesInc
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.
Φ² = Φ+1
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)
Squirrely
try mathway or photomath