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Calculate: ((cosπ/12+isinπ/12)(cosπ/24+isinπ/24))/(cosπ/8+isinπ/8)

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alexa0048 | Student, Grade 11 | eNoter

Posted March 18, 2013 at 3:14 PM via web

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Calculate:

((cosπ/12+isinπ/12)(cosπ/24+isinπ/24))/(cosπ/8+isinπ/8)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 18, 2013 at 4:09 PM (Answer #1)

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You need to use the following rules when performing the multiplication of division of two complex numbers, such that:

`(cos alpha_1 + i*sin alpha_1)*(cos alpha_2 + i*sin alpha_2) = cos(alpha_1 + alpha_2) + i*sin(alpha_1 + alpha_2)`

`(cos alpha_1 + i*sin alpha_1)/(cos alpha_2 + i*sin alpha_2) = cos(alpha_1 - alpha_2) + i*sin(alpha_1 - alpha_2)`

Reasoning by analogy yields:

`(cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)) = cos(pi/12 + pi/24) + i*sin(pi/12 + pi/24)`

`(cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)) = cos((2pi)/24 + pi/24) + i*sin((2pi)/24 + pi/24)`

`(cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)) = cos((3pi)/24) + i*sin((3pi)/24) `

`(cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)) = cos(pi/8) + i*sin(pi/8)`

Substituting `cos(pi/8) + i*sin(pi/8)` for `(cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24))`   yields:

`((cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)))/(cos(pi/8) + i*sin(pi/8)) = (cos(pi/8) + i*sin(pi/8))/(cos(pi/8) + i*sin(pi/8))`

Reducing duplicate factors yields:

`((cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)))/(cos(pi/8) + i*sin(pi/8)) = 1`

Hence, performing the operations of multiplication and division of the given complex numbers yields `((cos (pi/12) +i*sin(pi/12))(cos (pi/24) + i*sin(pi/24)))/(cos(pi/8) + i*sin(pi/8)) = 1.`

 

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