# Taking a number to a complex number power

1. Dec 16, 2011

How do you compute the following?

2it where t is a real number

while I am at it, how do you compute powers that are not integers

ie: 23.14

2. Dec 16, 2011

### The1337gamer

2^(it) = exp(it(log2)) = exp(i(tlog2)) = cos(tlog2) + isin(tlog2)

I guess we would define

2^(3.14) = exp(3.14(log2))

I'm not sure if i could write it in any other way, and i doubt i could compute that in my head but they are equivalent.

Sorry for the bad formatting if it isn't clear.

Edit: These logs are base e.

Last edited: Dec 16, 2011
3. Dec 16, 2011

### HallsofIvy

"3.14", as opposed to $\pi$, is a rational number. It is, in fact, $314/100= 157/50$ so $2^{3.14}= 2^{157/50}= \sqrt[50]{2^{157}}$.

$\pi$ is not rational but there exist a sequence of rational numbers that converge to it (the sequence 3, 3.1, 3.14, ..., for example). $2^\pi$ is equal to the limit of the sequence $2^3$, $2^{3.1}= 2^{31/10}= \sqrt[10]{2^{31}}=$, $2^{3.14}= 2^{314/100}= \sqrt[50]{2^{157}}$,...

Of course, to actually calculate those things you would use $2^a= e^{a ln(2)}$ as The1337gamer said.

4. Dec 17, 2011