# 1^i = e^(-2pi)?

#### devire

Registered Member
got another complex analysis question for u guys and girls here. would 1^i = e^(-2pi), since e^(2pi*i) = 1, making 1^i = e^(2pi*i*i) = e^(-2pi) = 1 / e^(2pi). i tried it with my TI-83 calculator and it just gave me 1. so, which is it?

oh and "pi" = 3.14..., i don't know how to make the symbol so i just say "pi".

edit: i know this is probably wrong, but since 1 = e^0 = e^(2pi*i), wouldn't that make 2pi*i == 0, making 0 * i == -2pi?

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devire said:
since e^(2pi*i) = 1
I haven't studies math for forty years, but . . .
doesn't e^(n.pi*i) = 1 where n is ANY even integer. So why pick on n = 2?
No power of 1 can ever deviate from unity, can it?

EEEk!

The thing is, you can rewrite a power using exponents and logs:
$$1^i = (e^{\log 1})^i = e^{i \log 1}$$
However, as River Ape pointed out, $$\log 1$$ isn't well defined in complex numbers. $$2 \pi i k$$ works $$\forall k \in Z$$. So there are infinitely many answers:
$$1^i = e^{i \log 1} = e^{i 2 \pi i k} = e^{-2 \pi k}$$

http://en.wikipedia.org/wiki/Complex_logarithm#Log.28z.29_as_a_multi-valued_function

You can turn log into a single valued function by putting an arbitrary branch cut somewhere (e.g. force the imaginary part to be $$\in (- \pi, \pi]$$ by convention) which is probably what your calculator did.

TI calculators (the only ones I've ever used) always give values on the principle branch of the function.

A similar problem arises when you try to compute arctangents. This is something the undergrads in my class never understand completely.

I haven't studies math for forty years, but . . .
doesn't e^(n.pi*i) = 1 where n is ANY even integer. So why pick on n = 2?
No power of 1 can ever deviate from unity, can it?

No, e^(n*pi*i) is 1 for n even, and -1 for n odd. Each increment of pi takes you half way around the unit circle.

I haven't studies math for forty years, but . . .
doesn't e^(n.pi*i) = 1 where n is ANY even integer. So why pick on n = 2?
No power of 1 can ever deviate from unity, can it?

for 1 and -1 raised to real powers, yes. but not when 1 or -1 is raised to a complex number, for example, (-1)^(-i) = e^pi.

P.S. why did u stop studying math for so long? u probably studied math pretty well if u could remember that after 40 years.

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I haven't studies math for forty years, but . . .
doesn't e^(n.pi*i) = 1 where n is ANY even integer. So why pick on n = 2?
No power of 1 can ever deviate from unity, can it?

Just to pick on you a bit more, the roots of unity are a very important mathematical concept. An n[sup]th[/sup] root of unity is a complex number that satisfies $$z^n=1$$. Raising both sides to the 1/n power yields z=1[sup](1/n)[/sup], hence the name "root of unity". For more, see http://en.wikipedia.org/wiki/Root_of_unity and http://mathworld.wolfram.com/RootofUnity.html as starters.

edit: i know this is probably wrong, but since 1 = e^0 = e^(2pi*i), wouldn't that make 2pi*i == 0, making 0 * i == -2pi?

They way I see it, the only way 2pi*i can be o is when i = o :shrug: