what is value of i^i where i is square root of minus one? is answer a real no?

Answer 1

Since i = cosπ/2 + isin(π/2)

= e^(iπ/2)

then

i^i = (e^(iπ/2))^i

= e^(i²π/2)

= e^(-π/2) (which certainly is a real number!!)

(≈ 0.20788)

To extend this notion even further:

In fact i = cos (π/2 + 2nπ) + isin(π/2 + 2nπ) where n = any integer

= e^i(π/2 + 2nπ)

Thus i^i = (e^i(π/2 + 2nπ))^i

= e^(i²(π/2 + 2nπ))

= e^(-(π/2 + 2nπ))

= e^(-(4n + 1)π/2) where n is any integer

So i^i = ....., e^(7π/2), e^(3π/2), e^(-π/2), e^(-5π/2), .....

(≈ ....., 59609.74, 111.32, 0.20788, 0.0003882 .......)

ie it is multivalued with an infinite number of values and with all its values in geometric progression BOTH ways from e^(-π/2) (≈ 0.20788) with a common ratio of e^(2π) (≈ 535.4917)

Answer 2

This question is not as straightforward as it looks because the complex functions involved are multi-valued, and the problem of selecting a continuous branch arises.
First of all, z^w must be defined, and this can be done by

z^w := exp(w ln(z))

which raises the question of what exactly is ln(z). Since z = |z|exp(i arg(z)), the log can be defined by

ln(z) := ln|z| + i arg(z)

(where ln|z| is the usual real log of the modulus). This reduces the question to choosing a continuous branch of the argument function. This cannot be done on the entire complex plane, but it can be done after removing any half-line that starts at the origin. The "usual" choice is to remove the negative real half-line, that is, to give up on defining the complex log of negative numbers. Then by arg(z) we mean the value of the argument that lies in (-Pi, Pi), often called the principal argument. Then z^w is defined and continuous (and actually holomorphic) for every complex w and z, except for a zero or negative z. In particular

i^i = exp(i ln(i)) = exp(i*iPi/2) = exp(-Pi/2)

which is real.


Note: Instead of giving up on defining the complex logs of negative numbers, we could have equally well given up on the positive numbers (unusual though this would be), by taking a branch of arg in (0, 2Pi). Then the complex logs of negative numbers would have been well defined, e.g. logarithm(-1) = iPi.

Answer 3

Absolutely!!!
if the square root of -1 is i, then i^i is most definately -1 which is a real number.

Answer 4

Wal C is absolutely correct.