Express 2+6i√3÷3 +i√3 in polar form?

1 Answer
Feb 25, 2018

(10.583*(cos(79.10°)+i*sin(79.10°)))/(3.464*(cos(30°)+i*sin(30°)))

Explanation:

First let's state the conversion between cartesian form and polar form

z=a+bi
z = r*(cos(theta)+i*sin(theta))
Where r = sqrt(a^2+b^2) and theta = arctan(b/a)

Let's break the equation into two smaller equations

The first: z_1 = 2+i*6sqrt(3)

r_1 = sqrt(2^2+(6sqrt(3))^2) = 10.583
theta_1 = arctan((6sqrt(3))/2) = 79.10°

And the second: z_2 = 3 +i*sqrt(3)

r_2 = sqrt(3^2+sqrt(3)^2) = 3.464
theta_2 = arctan(sqrt(3)/3) = 30°

Thus we can make the equation:

(10.583*(cos(79.10°)+i*sin(79.10°)))/(3.464*(cos(30°)+i*sin(30°)))