on i^i

[richardt]

Greetings: I am told that i^i has value e^-0.5pi. However in calculating this I begin with the fact that i = e^[(4k+1)pi*i/2] for all integers, k. Hence it seems that i^i = e^-[(4k+1)pi/2]. But this implies the expression has infinitely many real values - which is quite troubling to me.

Where am I going wrong here?

Thank you kindly,

Rich B.

 

[pka]

Greetings: I am told that i^i has value e^-0.5pi. However in calculating this I begin with the fact that i = e^[(4k+1)pi*i/2] for all integers, k. Hence it seems that i^i = e^-[(4k+1)pi/2]. But this implies the expression has infinitely many real values - which is quite troubling to me. Where am I going wrong here?

If each of \(\displaystyle z~\&~w\) is a complex number then \(\displaystyle z^w=\exp(w\log(z))\).

Now \(\displaystyle \log(i)=\ln(|i|)+i\left(\frac{\pi}{2}+2k\pi\right)=i\left(\frac{\pi}{2}+2k\pi\right)\)

So what is \(\displaystyle i~^i=~?\)

 

[Ishuda]

Greetings: I am told that i^i has value e^-0.5pi. However in calculating this I begin with the fact that i = e^[(4k+1)pi*i/2] for all integers, k. Hence it seems that i^i = e^-[(4k+1)pi/2]. But this implies the expression has infinitely many real values - which is quite troubling to me.

Where am I going wrong here?

Thank you kindly,

Rich B.

You aren't going wrong. As a general solution, in the same context i (or any other number) has infinitely many values. Not only is iⁱ = \(\displaystyle e^{-(2k+\frac{1}{2})\pi}\), it is also \(\displaystyle e^{(-(2k+\frac{1}{2})+2ni)\pi}\) so that the magnitude is always \(\displaystyle e^{-(2k+\frac{1}{2})\pi}\) but the argument is \(\displaystyle 2n\pi\). You will always, in that context, need to pick the Riemann sheet (and cut) you are working on which will determine the k (when you make the first choice of a sheet and cut) and n (when you make the second choice).

 

[richardt]

Thank you, folks