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B
e−π/2
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C
x
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D
1
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Solution
The correct option is Ae−π/2 xx=ii
Let z=ii
Taking log on both sides logz=ilogi =i[12log(02+12)+itan−110] =i[iπ2]=−π2 Z=e−π2
Method - II: xx=ii=(eiπ2)i=e−π2