# Thread: Converting from Polar Form to Complex Form?

1. ## Converting from Polar Form to Complex Form?

I understand the process of switching back to Complex form. However, there's one part that I don't understand where the answer comes from. Example:
Express 2(cos 2pi/3 + i sin 2pi/3) in rectangular form. The polar coordinate is (2, 2pi/3). That I understand.
Now, when you start to simplify the problem, it comes out to be;
2(-1/2 + i sqrt3/2) "sqrt" meaning square root of, as in the square root of 3.
I get where the negative one half came from. But the square root of three over two is where I'm lost. When I put into my calculator, "sin(2pi/3)" I get a really long decimal. Eventually, the answer comes out to be;
-1 + isqrt3.
I understand how they got -1. But if someone could PLEASE explain the square root of three over two I'd really appreciate it. Thank you!

Thanks!
Joe

3. ## Re: Converting from Polar Form to Complex Form?

Originally Posted by jtkjackie
I understand the process of switching back to Complex form. However, there's one part that I don't understand where the answer comes from. Example:
Express 2(cos 2pi/3 + i sin 2pi/3) in rectangular form. The polar coordinate is (2, 2pi/3). That I understand.
Now, when you start to simplify the problem, it comes out to be;
2(-1/2 + i sqrt3/2) "sqrt" meaning square root of, as in the square root of 3.
I get where the negative one half came from. But the square root of three over two is where I'm lost. When I put into my calculator, "sin(2pi/3)" I get a really long decimal. Eventually, the answer comes out to be;
-1 + isqrt3.
I understand how they got -1. But if someone could PLEASE explain the square root of three over two I'd really appreciate it. Thank you!
Can you answer the following questions?
1. $\sin \left( \frac{\pi}{3}\right) = ?$
2. $\sin \left( \pi - \theta) = ?$
3. $\sin \left( \frac{2\pi}{3}\right) = \sin \left( \pi - \frac{\pi}{3}\right) = ?$

4. ## Re: Converting from Polar Form to Complex Form?

draw trig diagram all stations to central and work out which quadrant its in starting from imaginary axis you move counter clockwise. Also draw the 30 60 triangle with corresponding sides. This may help you. The 2 is the modulus gained from the square of the other sides of the triangle. |r| is the modulus or hypotenuse and since it is 2 the other sides are sqrt3 and 1.