Question

How do you write the following in trigonometric form and perform the operation given `(2+2i)(1-i)`?

Answer 1

`(2+2i)(1-i)=4`

Explanation:

Let `z_1=2+2i` And `z_2=1-i`
Then the moduli are `∣z_1∣ = sqrt (2^2+2^2)=2sqrt2`
and `z_2= sqrt(1+1)=sqrt2`
Rewrite the numbers
`z_1=2+2i=2sqrt2(1/sqrt2+i/sqrt2)=2sqrt2(costheta_1+isintheta_1)`
Here `theta_1=pi/4`
so `z_1=2sqrt2(cos(pi/4)+isin(pi/4))= 2sqrt2e^(ipi/4)`
Similarly
`z_2=1-i=sqrt2(1/sqrt2-i/sqrt2)=sqrt2(costheta_2-isintheta_2)`
`theta_2=-pi/4`

so `z_2=sqrt2(cos(-pi/4)+isin(-pi/4))= sqrt2e^(-ipi/4)`
so the result is `z_1*z_2=2sqrt2e^(ipi/4)*sqrt2e^(-ipi/4)=4`