Stuck on complex problem! Find the product of the complex numbers. Express your answer in trigonometric form. z1 = 3 (cos(pi/4) + i sin(pi/4)) z2 = 7 (cos(3pi/8) + i sin(3pi/8)) A. 3/7 (cos(5pi/8) + i sin(5pi/8)) B. 3/7 (cos(-pi/8) + i sin(-pi/8)) C. 21(cos(5pi/8) + i sin(5pi/8)) D. 21(cos(-pi/8) + i sin(-pi/8))

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Stuck on complex problem! Find the product of the complex numbers. Express your answer in trigonometric form. z1 = 3 (cos(pi/4) + i sin(pi/4)) z2 = 7 (cos(3pi/8) + i sin(3pi/8)) A. 3/7 (cos(5pi/8) + i sin(5pi/8)) B. 3/7 (cos(-pi/8) + i sin(-pi/8)) C. 21(cos(5pi/8) + i sin(5pi/8)) D. 21(cos(-pi/8) + i sin(-pi/8))

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I am thinking that the answer would be C. Having some trouble however.
\[z _{1}=r _{1} \left( \cos \theta1+ \iota \sin \theta1 \right)=r _{1}e ^{i \theta1}\] \[z _{2}=r2\left( \cos \theta2+\iota \sin \theta2 \right)=r2e ^{i \theta2}\] \[z _{1}*z _{2}=r1*r2e ^{i \left( \theta1+\theta2 \right)}\] \[=r1*r2\left( \cos \left( \theta1+\theta2 \right)+\iota \sin \left( \theta1+\theta2 \right) \right)\]
Oh ok! So from looking at the formula, I can see right away that the answer would not be A or B because we multiply 3 by 7.

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Other answers:

correct
Now we have to add pi/4 and 3pi/8 which gives us cos(5pi/8) which is C! Thank you!!
well done
I wish I knew about this formula sooner hahaha! Thank you so much!
yw

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