How do I power a number to i?

Answer 1

Just do it.
Multiply it by itself i times.

Oh you meant i = sqrt(-1).
You need to clarify that man.

a^i = exp(i*lna)
= cos(lna) + i*sin(lna)

Answer 2

I is the square root of negative 1, thusly it is an imaginary number. It is more likely that you are being asked to raise i to the power of another number.

For example, i to the power of 2 = -1

See link below for more help

Answer 3

are you asking i to a power? as in i, i^2, i^3, i^4 etc... if so there is a pattern

i^1= i
i^2= -1
i^3= -i
i^4= 1

and this pattern keeps repeating forever... so i^5 would be i again

the reason for this is i = sqrt of -1 so i^1 would just be sqrt of (-1)

i^2 would be sqrt of (-1) * sqrt of (-1) leaving you with -1...

Answer 4

You mean, how do you calculate something like 2^i? In this case, the fact that x = e^ln(x) is helpful, along with Euler's formula, e^(iθ) = cos(θ) + isin(θ).

So let's say we want to find x^i, for some real number x. This is:

x^i =
e^(ln x^i) =
e^[i ln(x)] =
cos(ln(x)) + i sin(ln(x))

Answer 5

Someone else has already suggested, correctly, Euler's formula, which is e^ix = cos x + i * sin x

So, if you have a^i, it's e^(i * ln(a)) = cos(ln(a)) + i * sin(ln(a))

Answer 6

a^i = e^(i * ln(a))

use Euler's relation to solve the e^(i*ln(a)) part
(example: e^(pi*i) = -1)

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