How do I solve by de moivres theorem and in

Carla Murphy

Carla Murphy

Answered question

2022-01-17

How do I solve by de moivres theorem and in a final answer of polar form ((cis2θ)3)((cis5θ)2(cis3θ)4((5ej2θ)3)?

Answer & Explanation

vicki331g8

vicki331g8

Beginner2022-01-18Added 37 answers

(cis2θ)3(cis5θ)2(cis3θ)4(5ej2θ)3
=(cis(6)θ)(cis10θ)(cis12θ)(5cis2θ)3
=(cis4θ)(cis12θ)(125cis6θ)
=cis4θ(cis12θ)(125cis6θ)
=1125cis(14)θ
poleglit3

poleglit3

Beginner2022-01-19Added 32 answers

I guess that the letter j in ej2θ stands for the complex number i and therefore
(5ej2θ)3=(5cis(2θ))3
Now, from de Moivre identity and the rules of the exponents we get
(cis(2θ))3=(cis(θ))6,(cis(5θ))2=(cis(θ))10
(cis(3θ))4=(cis(θ))12,(5ei2θ)3=53(cis(θ))6
Answer: ((cis(2θ))3)((cis(5θ))2)((cis(3θ))4)((5ei2θ)3)=53(cis(θ))(6+10126)=153(cis(θ))14=cis(14θ)53
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

De Moivre's theorem states that (cosθ+isinθ)n=cos(nθ)+isin(nθ) - so you can rewrite the expression above as:(((cosθ+isinθ)2)3)(((cosθ+isinθ)5)2)(((cosθ+isinθ)3)4)((5(cosθ+isinθ)2)3)=((cosθ+isinθ)6)((cosθ+isinθ)10)((cosθ+isinθ)12)(125(cosθ+isinθ)6)=1125(cosθ+isinθ)14=1125e14iθ

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