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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