# what is i(root over -1) to the power i?what type of complex no it will be?and pls tell me something about logarithm of complex numbers

Grade:12

## 2 Answers

Chetan Mandayam Nayakar
312 Points
12 years ago

MY PREVIOUS SOLUTION WAS WRONG.

(√(-1))i=e(i(((pi/2)+2n*pi)))*i, where n is any integer. Therefore, there are infinitely many values.

jitender lakhanpal
62 Points
12 years ago

dear Debanuj

i to the power i = e-∏/2

where

e= 2.7108   and    ∏ = 22/7

proof

let

y = i to the power i

taking natural logarithms on both sides (ln is log base e)

ln y = i ln i

changing i into eulerian form

i = e to the power i∏/2

ln y = i*i∏/2

as i*i = -1

lny = -∏/2

y = e to the power -∏/2

so proved

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