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Question

Which of the following is a fourth root of 12+i32?


A

cisπ12

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B

cisπ2

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C

cisπ6

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D

cisπ3

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Solution

The correct option is A

cisπ12


Explanation for the correct option:

Step 1: Change the equation into polar form

Complex equation Z=a+ib

Let Z=12+i32 ….(1)

Therefore, a=12 and b=32

Polar form z=r(cosθ+isinθ) ….(2)

r=a2+b2

r=(12)2+(32)2r=14+34r=1+34r=44r=1r=1

Now

cosθ=arcosθ=12cosθ=cosπ3

and

sinθ=brsinθ=32sinθ=sinπ3

Equate equation(1) and equation(2),

12+i32=r(cosθ+isinθ)12+i32=cosπ3+isinπ3

Step 2: Write the equation in exponential form

The exponential form of a complex number cosθ+isinθ is eiθ.

cosπ3+isinπ3=eiπ3

The fourth root of the equation is,

eiπ34=eiπ12eiπ12=cosπ12+isinπ12

Step 3: Write the equation in cis from

We know that, cosx+isinx=cisx

cosπ12+isinπ12=cisπ12

Hence, the correct option is option (a).


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