# Does $1^i$ and $1^{\frac{0}{0}}$ also give $1$ again? [duplicate]

This is the property of Real number $1$ that,

$1^n=1$ does this property only hold $\forall n \in \mathbb R$ or also $1^i=1$ and $1^{\frac{0}{0}}=1$

If it is; explain how?

I think that it should not give us $1$ again as we can't determine what is the square root of number $-1$ and similarly $\frac{0}{0}$ is indeterminate and so we can't proceed. May be I am wrong.

## marked as duplicate by Sufyan Naeem, Bumblebee, Jonathan Y., egreg, Cameron BuieApr 30 '15 at 10:26

• $1^i=\left(e^0\right)^i=e^{0\cdot i}=e^0=1$ – Alexey Burdin Apr 30 '15 at 9:42
• As for $1^i$: this answer to a related question says it all: math.stackexchange.com/a/3674/101420 – Vincent Apr 30 '15 at 9:55
• My question involves "what is $1^{\frac{0}{0}}$" and the question you are talking about doesn't. – Sufyan Naeem Apr 30 '15 at 9:59
• But $\frac00$ isn't a number, so what would it even mean to talk about $1^\frac00$? – Cameron Buie Apr 30 '15 at 10:28
• $\frac{0}{0}$ is indeterminate that is, we don't know that how many $0s$ are in $0$. Either one or many zeros or no zero. So I think that $1^{\frac{0}{0}}$ may or may not give us $1$. That confused me. – Sufyan Naeem Apr 30 '15 at 11:07

$1^i$ is defined as $e^{i \ln 1}$

$\ln 1$, though, it's not our real logarithm, but it's the complex one; the point is that is a multi-valued function; to be more precise,

$$\ln z = \ln_\mathbb R |z| + (\arg z + 2k\pi)i$$

Where with $\ln_\mathbb R$ I mean the usual real logarithm.

So $\ln 1 = \ln_\mathbb R 1 + (2k\pi)i = 2k\pi i$

Hence $$1^i = e^{i \ln 1} = e^{-2k\pi}$$

So $1^i$ it's not really a number, it's more like a set of numbers; all the numbers in the form $e^{-2k\pi}$, $k \in \mathbb Z$.

The value $k=0$ is special and it's sometimes called the principal value; choosing this value for $k$ clearly results in $1^i = 1$, but it's important to understand that $1^i$ is not a well-defined number.

Note that, as @Jonathan Y. points out, a similar argument also holds for $1^{3/5}$. And if one thinks about it also for $\sqrt 1$, which is not a number but the set $\pm 1$.

The point is that we can define without ambiguity $\sqrt 1$ (take the positive value) and $1^{3/5}$ (there is only one real root).

One could naively think that the same could be done with $1^i$; just define in some way a special value and work with that. The point is that it can't be done; there is no way to specify a value without ambiguity and carry on all the usual algebra rules. So we're stuck with this multi valued definition if we are in the complex plane; if we are on the real line, instead, it's possible to define $a^{b/c}$ as a single value.

P.S. The principal value I referred to earlier it's sometimes useful but problems arise if you define the complex logarithm as it's principal value. Once you start studying complex analysis and (for example) the residue theorem it should be clear why :-)

• A similar argument could be made for, e.g., $1^{3/5}$, which (I think) is even more startling. – Jonathan Y. Apr 30 '15 at 10:10
• @JonathanY. Thanks! I've edited the answer to include it :) – Ant Apr 30 '15 at 10:19

$1^i$ is a multi-valued quantity $$e^{i\theta}=\cos\theta+i\sin\theta\implies e^{2n\pi i}=1\implies1^i=\color{RED}{e^{-2n\pi }}$$ for any integer $n.$

Maybe you would like to know that the exponentiation by complex number does not work very well. The previous answers are correct, but what if you think like this?

$$1^i=(i^4)^i=(i^i)^4=((e^{i\frac{\pi}{2}})^i)^4=(e^{-\frac{\pi}{2}})^4=e^{-2\pi} \neq 1$$

Where is our God now? You have to be very careful when you do exponentiation with complex numbers.