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

1 Answer
Feb 14, 2016

#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).