Question

Write z = 9 - 3 sqrt 3i in polar form.

Thank you so much!!!

Answer 1

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))

Answer 2

z = 6√3(cos(11π/6)+isin(11π/6))

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.