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\( 1^{i} = ? \)

What is \( 1^{i} \) ?

We can say that \( e^{2\pi i} = 1 \), so then \( (e^{2\pi i})^{i} = e^{-2\pi }= 1^{i} \) and \( e^{-2\pi} \approx 0.001867 \). However, we can also say that \( e^{0} = 1 \), so then \( (e^{0})^i = e^0 = 1 \), so then \( 1^{i} = 1 \). And you can use \( 4 \pi, 6 \pi, 8\pi, 2n \pi, n \in \mathbb{Z} \), because the domain of Euler's formula \( e^{ix} = \cos x + i \sin x \) is \( \mathbb{R} \). Google and Wolfram Alpha both say that the answer is \( 1 \), so how do we know that the answer is \( 1 \)? 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
6 months, 4 weeks ago

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Complex exponents are multi-valued.

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

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

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

Calvin Lin Staff - 6 months, 3 weeks ago

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\( \dfrac{\sqrt{1}}{\sqrt{-1}} \neq \sqrt{\dfrac{1}{-1}} \) because for the property \( \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}} \), \( a, b \in \mathbb{P} \). Since we define complex numbers given as coordinates \( (a, b) \) where \( a, b \in \mathbb{R} \), a complex valued function is a mapping from \( \mathbb{R} \to \mathbb{C} \). By definition, two complex numbers \( (a, b) \) and \( (c, d) \) are equal iff \( a= c \wedge b=d \), we write \( i \) as \( (0, 1) \) and \( -i \) as \( (0, -1) \), 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 \( e^{i \theta } = \cos \theta + i \sin \theta \) allows for an infinite number of \( \theta \) that can return the same complex numbers.

Hobart Pao - 6 months, 2 weeks 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 \( \sqrt{ ab} = \sqrt{a} \sqrt{b} \) (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 - 6 months, 2 weeks ago

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

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

Anirudh Sreekumar - 6 months, 4 weeks ago

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I mean I found a way to get \( 1 \) 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 - 6 months, 4 weeks ago

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

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

Anirudh Sreekumar - 6 months, 4 weeks ago

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

Hobart Pao - 6 months, 3 weeks 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 - 6 months, 3 weeks ago

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