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Question

Let fp(α)=eiαp2.e2iαp2.e3iαp2.e4iαp2eiαp, (where i=1 and pN) then limnfn(π) is


A

1

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B

- 1

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C

i

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D

- i

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Solution

The correct option is C

i


fp(α)=eiαp2(1+2+3+4++p)=eiαp2.p(p+1)2=eiα2(1+1p) limnfn(π)=limneiπ2(1+1n)=eiπ2(1+0)=eiπ2=cosπ2+i sinπ2=i


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