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i i

ii
 

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
7 years ago
Hello student,
Please find the answer to your question below
Use identity a = e^(lna).
i^i = e^[ln(i^i)] = e^[i*lni]
Complex logaritm lnz = ln|z| + i*argz
Then ln(i) = ln |i| + i*arg(i) = ln1 + i*pi/2 = i*pi/2.
So
i^i = e^[i*lni] = e^[ i*i*pi/2] = e^ [ -pi/2 ].
Note:pi above denotes\pi

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