How do you simplify 2(cos(pi/2)isin(pi/2))*2(cos((3pi)/2)+isin((3pi)/2))2(cos(π2)isin(π2))2(cos(3π2)+isin(3π2)) and express the result in rectangular form?

1 Answer
Sep 11, 2017

The answer is 4.

Explanation:

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

z*w = (2*cis(pi/2))*(2*cis((3pi)/2))zw=(2cis(π2))(2cis(3π2))

The product rule for polar form states that for two numbers, z_1=r_1*cis(theta_1)z1=r1cis(θ1) and z_2=r_2*cis(theta_2)z2=r2cis(θ2), you have z_1*z_2=r_1*r_2*cis(theta_1 + theta_2)z1z2=r1r2cis(θ1+θ2).

Doing this process, we get the following:

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

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

z*w = 4zw=4.