Respuesta :
z=a+bi =9-(3√3)i , a=9, b=-3√3
You can draw this point on the x/y plane, where a (real part) is x-coordinate, and b (imaginary part) is y-coordinate.
z=r(cosα +i*sinα)
From triangle ABC we can see, that
r=√(a²+b²)=√(9²+(-3√3)²)=√(81+27)=3√(9+3)=3√12
tan(α)=(-3√3)/9=-(√3)/3
α=tan⁻¹(-(√3)/3)=-π/6
But we have to write an angle as positive value, so
α = 2π-π/6=12π/6-π/6=11π/6
So, z=r(cosα +i*sinα) = 3√12(cos(11π/6) +i*sin(11π/6))
You can draw this point on the x/y plane, where a (real part) is x-coordinate, and b (imaginary part) is y-coordinate.
z=r(cosα +i*sinα)
From triangle ABC we can see, that
r=√(a²+b²)=√(9²+(-3√3)²)=√(81+27)=3√(9+3)=3√12
tan(α)=(-3√3)/9=-(√3)/3
α=tan⁻¹(-(√3)/3)=-π/6
But we have to write an angle as positive value, so
α = 2π-π/6=12π/6-π/6=11π/6
So, z=r(cosα +i*sinα) = 3√12(cos(11π/6) +i*sin(11π/6))

Answer:
z = 6√3(cos(11π/6)+isin(11π/6))
Step-by-step explanation:
We have given an equation in complex form.
z = 9-3√3i
z = a+bi is general form of comple number.
on comparing above two equations, we get
a = 9 and b = -3√3
We have to write it in polar form.
The formula to convert an equation in ploar form is:
z = r(cos∅+isin∅) where r is magnitude of z and ∅ is angle.
r = √a²+b² and ∅ = tan⁻¹(b / a)
putting the values of a and b in above two formulas , we have
r = √(9)²+(-3√3)² = √81+27 = √108 = √36×3 = 6√3
∅ = tan⁻¹(-3√3 / 9)
∅ = tan⁻¹(-1/√3)
∅ = -π / 6
we have to change ∅ in positive.
∅ = 2π-π/6 = 12π-π / 6
∅ = 11π / 6
putting the value of r and ∅ in given formula, we get
z = 6√3(cos(11π/6)+isin(11π/6)) is the polar form of given complex number.