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

1 Answer
Narad T.
Oct 20, 2016

(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

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