Question

How do you plot `9sqrt{3} + 9i` on the complex plane and write it in polar form?

Answer 1

`18cos(pi/6)+ isin(pi/6)`

Explanation:

`rho=sqrt((9sqrt(3))^2+9^2)=sqrt(81*3+81)=sqrt(324)=18`

`theta=arctan(9/(9sqrt3))=sqrt(3)/3`

the number is in the form a+bi, in which a and b are positive, therefore, the arc that matters here is `pi/6`

So the number in polar coordinates is:

`18cos(pi/6)+ isin(pi/6) or 18e^(pii/6)`

The drawing in the complex plane is a triangle in the first quadrant,
with x=9 and y=9sqrt(3).