Thread: Polar to Complex Form

1. Polar 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!

2. Re: Polar 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!
Calculators are of absolutely no use when it comes to this topic.
One just has to know the special values for $\displaystyle 0,~\frac{\pi}{6},~\frac{\pi}{4},~\frac{\pi}{3},~ \frac{\pi}{2},~\frac{2\pi}{3},~\frac{3\pi}{4},~ \frac{5\pi}{6}~\&~ \pi$.

So learn them.

3. Re: Polar 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;
Was that "really long decimal" 2.0943951023931954923084289221863? Probably you have your calculator in "degree
mode" when it should be in radians.
In any case, as Plato says, you really should know some 'basic" values of sine and cosine.

-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!