1i=? 1^{i} = ?

What is 1i 1^{i} ?

We can say that e2πi=1 e^{2\pi i} = 1 , so then (e2πi)i=e2π=1i (e^{2\pi i})^{i} = e^{-2\pi }= 1^{i} and e2π0.001867 e^{-2\pi} \approx 0.001867 . However, we can also say that e0=1 e^{0} = 1 , so then (e0)i=e0=1 (e^{0})^i = e^0 = 1 , so then 1i=1 1^{i} = 1 . And you can use 4π,6π,8π,2nπ,nZ 4 \pi, 6 \pi, 8\pi, 2n \pi, n \in \mathbb{Z} , because the domain of Euler's formula eix=cosx+isinx e^{ix} = \cos x + i \sin x is R \mathbb{R} . Google and Wolfram Alpha both say that the answer is 1 1 , so how do we know that the answer is 1 1 ? Is there some domain restriction I'm not aware of? Or maybe it's because we raised an expression to a complex exponent.

Note by Hobart Pao
2 years, 6 months ago

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Complex exponents are multi-valued.

So, when we say 1i=1 1 ^ i = 1 , what it means is that "One of the many values of 1i 1^i is equal to 1". Note that the equal sign doesn't have the transitive property, since it is true that e2π=1i=1 e ^ { - 2 \pi } = 1 ^ i = 1 .

This is how various misconceptions/errors in complex numbers can arise. The standard example is looking at i=11=11=1=11=11=i. -i = \frac{ \sqrt{1} } { \sqrt{ -1 } } = \sqrt{ \frac{ 1 } { -1 } } = \sqrt{ -1 } = \sqrt{ \frac{ -1} { 1 } } = \frac{ \sqrt{ -1} } { \sqrt{ 1 } } = i .

Using the language of multi-value functions, do you see how to resolve the apparent contradiction?

Calvin Lin Staff - 2 years, 5 months ago

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1111 \dfrac{\sqrt{1}}{\sqrt{-1}} \neq \sqrt{\dfrac{1}{-1}} because for the property ab=ab \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} , a,bP a, b \in \mathbb{P} . Since we define complex numbers given as coordinates (a,b) (a, b) where a,bR a, b \in \mathbb{R} , a complex valued function is a mapping from RC \mathbb{R} \to \mathbb{C} . By definition, two complex numbers (a,b) (a, b) and (c,d) (c, d) are equal iff a=cb=d a= c \wedge b=d , we write i i as (0,1) (0, 1) and i -i as (0,1) (0, -1) , so they are not equal. Also, the previously mentioned property with the square roots applies to what was done because the domain for complex valued function is real numbers. This definition also allows for the multiple values you mentioned since eiθ=cosθ+isinθ e^{i \theta } = \cos \theta + i \sin \theta allows for an infinite number of θ \theta that can return the same complex numbers.

Hobart Pao - 2 years, 5 months ago

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With multi valued functions, you have to choose which interpretation you want, and then stick with it throughout your thinking. This does impact how we draw conclusions, but shouldn't lead to any errors when done properly.

Personally, I prefer to take the interpretation of "equality means that the intersection of these 2 sets is non-empty", we can preserve ab=ab \sqrt{ ab} = \sqrt{a} \sqrt{b} (where these roots are multi-valued, and not single-valued). However, the downside as mentioned is that equality isn't transitive (and so not a great symbol for people to manipulate naively).

Note that if we take "equality means that these 2 sets are identical", then it would hard to have any meaningful statements.

Calvin Lin Staff - 2 years, 5 months ago

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I'm not too sure about the following answer,but it was what i could think of immediately.

1i=(11)i=1i1i=11^i=\left(\dfrac{1}{1}\right)^i=\dfrac{1^{i}}{1^{i}}=1

Anirudh Sreekumar - 2 years, 6 months ago

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I mean I found a way to get 1 1 too but I'm not sure why we can't discount those other possible answers. I suspect there's some domain restriction that I missed but couldn't find any yet.

Hobart Pao - 2 years, 6 months ago

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no, i mean if you set any other value for 1i1^i like e2πe^{-2\pi}

it would imply that e2π=1e^{-2\pi}=1 which is not true

Anirudh Sreekumar - 2 years, 6 months ago

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@Anirudh Sreekumar but e2πi=cos2π+isin2π=1 e^{ 2 \pi i} = \cos 2\pi + i \sin 2\pi = 1 , so why can't I raise e2πi e^{2 \pi i } to the i i exponent?

Hobart Pao - 2 years, 5 months ago

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@Hobart Pao So my guess is that the laws of exponents are only for the real number system and that they break down when we get to complex numbers. I haven't learnt complex analysis yet so maybe I'll have the answer when I get there.

Hobart Pao - 2 years, 5 months ago

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