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?

1 Answer
Sep 11, 2017

Answer:

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#.