# Re: i to the ith power

```It's all in the logarithm!. Once you know how to find the log and the
exponential of numbers, then the basic _definition_ of exponentiation a^b
is reduced to multiplication, as follows:

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

Therefore, i^i becomes  e^(i*ln(i)).

But, you might say, how do you find the logarithm of a complex number?
This gets mathematically interesting (which is to say, uninteresting to non-
mathematicians). A good, readable book on the history of i, which deals
specifically with the question of i^i, has been published in the last year.
Unfortunately, my copy is sitting at home at the moment -- perhaps someone
else on this list could provide the reference (or you could look it up on
Amazon). Suffice to say, deep waters here...

All this relates to the _theory_ of how it's done and what it means. How the
mathematicians and engineers at TI handled this in _practice_ is only known by
TI. Why should sqrt(-1) be treated differently from i as a keypress? As noted
above, the relevant point should be whether ln(i) and ln(sqrt(-1)), which
should be the same as ln(-1)/2, are calculated differently. You might check
the TI tech reference files on this point.

As a sidelight, this is another good example of the fact that, in today's
"calculator dependent" environment, a lot of fascinating, deep topics are
explorable by those curious enough to just start playing around, seeing what
they can do. How can this be a bad thing? Who wants to go back to the old days?

RWW Taylor
National Technical Institute for the Deaf
Rochester Institute of Technology
Rochester NY 14623

>>>> The plural of mongoose begins with p. <<<<

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Steve Kraisler writes:

> Why when I take (a)  sqroot(-1) ^sqroot(-1) or (b) i ^ sqroot(-1) or (c)
> sqroot(-1)^i  the ti-83 gives an undefined response (specifically an error
> message) but when I try i^i  it gives back .2078795764
>
> How does the ti-83 handle an imaginary raised to an imaginary power?
>

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