MATH SOLVE

4 months ago

Q:
# Write z = 9 - 3 sqrt 3i in polar form.Thank you so much!!!

Accepted Solution

A:

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))