Question

How do you simplify `2(cos(pi/2)isin(pi/2))*2(cos((3pi)/2)+isin((3pi)/2))` and express the result in rectangular form?

Answer 1

The answer is 4.

Explanation:

I prefer the `cis` notation for polar form, so if we write it like that, we get, calling the above expression `z*w`, this:

`z*w = (2*cis(pi/2))*(2*cis((3pi)/2))`

The product rule for polar form states that for two numbers, `z_1=r_1*cis(theta_1)` and `z_2=r_2*cis(theta_2)`, you have `z_1*z_2=r_1*r_2*cis(theta_1 + theta_2)`.

Doing this process, we get the following:

`z*w = 4*cis(2pi) -= 4*cis(0)`.

Since `cis(0) = cos(0) + i*sin(0)`, we get the final result:

`z*w = 4`.