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Question

If ex+iy=α+iβ, then x+iy is called logarithm of α+iβ to the base e.loge(x+iy)=loge(reiθ) =loger+iθ where r is modulus value of x+iy & θ be the argument of x+iy If iα+iβ)=α+iβ, then α2+β2 equals

A
e(2n+1)πα2
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B
eπβ
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C
e(4n+1)αβ
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D
None of these
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Solution

The correct option is C eπβ
i(α+iβ)=α+iβ
(α+iβ)logi=log(α+iβ)
log(α+iβ)=(α+iβ)[iπ2+2nπi],nϵI
12log(α2+β2)+itan1βα=i(α+iβ)(4n+1)π2
Equating imaginary parts we have
12log(α2+β2)=βπ2(4n+1) α2+β2=e(4n+1)πβ

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