Question

What is 1^i?

Answer 1

i think 1

Answer 2

Wolfram says 1
I think 1
However I have read somewhere that there could be infinite solutions or something...

Answer 3

i don't think why 1^(any number, even complex) is not 1

Answer 4

\(\huge 1^{a+ib}=1\)

Answer 5

Hmmmm okay thanks.
Do you know about Euler's formula?
The one where e^(ix)=cos(x)+isin(x)

Answer 6

lets try
let 1^i =a
i log 1 = log a
log a = 0, gives a=1

yes i know

Answer 7

Does it have a domain restriction?

Answer 8

1 to the power of any value, either complex or real, is always 1.

Answer 9

@apple_pi is 'i' here referring to \(\bf \sqrt{-1}\) or is it a variable?

Answer 10

Domain restriction for Euler's formula i meant

Answer 11

Sqrt(-1)

Answer 12

No there is no domain restriction for 'x' in euler's formula.

Answer 13

no domain restriction, because there's no restriction for values of x in sin x or cos x

Answer 14

It's all real numbers. You should be able to deduce that just by looking at the formula.

Answer 15

Ok so then what's stopping me from saying:
e^0=e^(2ipi)
ln(both sides)
Therefore 0=2ipi?

Answer 16

\[\bf e^{i(0)}=\cos(0)+isin(0) =1\]

Answer 17

@apple_pi is correct

Answer 18

Wait, I'm correct?

Answer 19

this is new to me,
what made u write log 1 = 2pi k +Log 1 ??

Answer 20

oh, from this?
\(z=\exp(2\pi k+\operatorname{Log}z)\)

Answer 21

@hartnn given a complex number \(z=re^{i\theta}\) we can take its logarithm \(\log z=\log r+\log e^{i\theta}=i\theta+\log r\) but observe \(z=re^{i\theta}=re^{i(\theta+2\pi k)}\) yes? hence \(\log z=i\theta+2\pi ik+\log r\) as well. hence the complex logarithm is not a function as it is multivalued

Answer 22

@hartnn yeah I meant \(2\pi ki\)

Answer 23

this is common with complex functions; to make a complex logarithm function, we use a "branch cut" and limit ourselves to arguments say \(\theta\in[0,2\pi)\) -- this we call the principal complex logarithm \(\Log z\)

Answer 24

it's the same principle we use when taking square roots; observe that a real number has two real square roots, \(\pm x\). it makes sense then that the square root is not a function since it's multivalued. to deal with this we employ a branch cut at \(x=0\) and choose the \(x>0\) half to be our principal square root \(\sqrt{x}\)

Answer 25

even closer, consider the trigonometric functions. they do not have normal inverses in the sense they'd be multivalued since our function is not a bijection; this forces us to make a branch cut and limit ourselves to angles in say \((-\pi,\pi]\), for example.

Answer 26

@oldrin.bataku what am I correct about?

Answer 27

@apple_pi there are indeed infinite solutions \(z=\exp(2\pi k)\)

Answer 28

Inifinite solutions to 1^i?

Answer 29

you meant 2pi *i*k again ?

Answer 30

well, solution is the wrong word -- but \(1^i\) is ambiguous. we define complex exponentiation to be \(z^w=\exp(\log z^w)=\exp(w\log z)=\exp(w(2\pi ki+\Log z))\). Here we have \(z=1\) hence \(\Log z=0\) and \(w=i\) so we get \(1^i=\exp(i(2\pi ki+0))=\exp(-2\pi k)\)

Answer 31

@hartnn nah I kept making typos. it should be \(1^i=\exp(-2\pi k)\) or equivalently \(1^i=\exp(2\pi k)\) for any integer \(k\)

Answer 32

ohh, okk

Answer 33

@hartnn you're correct \(1\) is one of the possible results, as are \(e^{2\pi}\) and \(e^{-100\pi}\)

Answer 34

yes, i could get that, thanks :)

Answer 35

\[1^i=1\]

Answer 36

@Zarkon ... read above please. You are only further confusing the asker.

Answer 37

I'm well aware of the above (I have taught complex analysis)

The answer depends on ones definition. Some definitions use the real valued log when the base is a positive real number and some definitions use what you have above. The OP should consult their book/instructor to see what definition is being used.

Answer 38

The real valued log? I'm guessing you mean the principal value of the logarithm.

Answer 39

obviously.

here is one definition

http://en.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases_2

Answer 40

Okay, well vaguely referring to a 'real valued log' was "obviously" not clear.

Answer 41

that being said, that was an interesting link. thanks.