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I am stuck on the idea $i^i$ equals $1 \over 5$ I have seen proof of it, but I think the explanations don't make sense, I understand the maths itself but what I don't understand is how you assume it is a number but it is the square root of $-1$ and thus is I but also undefined because no numbers squared achieve, therefore you can substitute undefined^undefined, and a number divided by zero is undefined so what we now get is that: ${({5 \over 0})^{5 \over 0}} = 0.2$ which isn't logical so does anyone knows proof of $i^i=0.2$ without assuming it is a number I would imagine you can't but please tell me why you can assume $i$ to be a number?

Ethan
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Areze
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  • https://math.stackexchange.com/questions/191572/prove-that-ii-is-a-real-number, https://math.stackexchange.com/q/2568146/42969 – Martin R Mar 16 '21 at 19:50
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    You don't assume that $i$ is a number because $i$ is a number - nothing to assume here. And with polar coordinates you find $$i^i = (e^{i \frac{\pi}{2}})^i = e^{- \frac{\pi}{2}}$$ and is has nothing to do with $\frac{1}{5}$ – Lukas Mar 16 '21 at 19:50
  • but it is impossible for any number to equal square root of -1 basic isn't it – Areze Mar 16 '21 at 19:51
  • @Lukas: except that $e^{-\pi/2}\approx\frac15$. –  Mar 16 '21 at 19:51
  • Ok, sorry. I didn't calculate that. That was a little lazy by me – Lukas Mar 16 '21 at 19:52
  • You seem to be rejecting the very existence of $i$ in the first place and seem to be conflating one type of "undefined" mathematical object with another. Your first step should not be worrying about what $i^i$ is equal to but rather to concern yourself with understanding what $i$ is in the first place and what the complex number field is. – JMoravitz Mar 16 '21 at 19:57
  • The short version is that $i$ is very much well defined and well understood by the mathematical community at large. You just seem to have not caught up to the rest of us yet. Maybe start by reading Do complex numbers really exist? and What are imaginary numbers?. – JMoravitz Mar 16 '21 at 19:59
  • does anyone have an answer which explains what @JMoravitz has suggested because i just want to learn – Areze Mar 16 '21 at 19:59
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    @Areze You have to understand complex exponentiation before understanding what the value of $i^i$ is. – David P Mar 16 '21 at 20:01
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    "Why you can assume $i$ to be a number?" Is $0$ a number? Is $-1$ a number? Is $\sqrt 2$ a number? Why? – Théophile Mar 16 '21 at 20:04
  • A better rational approximation is 122397/588788, which is accurate to 13 decimal places. Here are a few more digits: 0.2078795763507619085469556198349787700338778416317696080751 – PM 2Ring Mar 16 '21 at 20:40

1 Answers1

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It is a pretty long story.

It is true that there is no real number $r$ such that $r^2=-1$ hence $-1$ has no square root. We will not contradict that.

But it turns out that if you accept a special symbol denoted $i$ and form expressions like

$$r+is$$ where $r,s$ are two real numbers, you define a whole new arithmetic that has extraordinary and very useful properties.

We define the addition and multiplication as follows:

  • $(r+is)+(r'+is')=(r+r')+i(s+s')$,

  • $(r+is)(r'+is')=(rr'-ss')+i(rs'+r's)$.

(as arithmetic on the reals is known, these formulas extend the arithmetic to those special expressions).

In particular, $i^2=i\cdot i=(0+i\cdot1)(0+i\cdot1)=-1+i\cdot0=-1$. There is no contradiction with what is written above, as $i$ is not a real number, just a conventional symbol.

From these definitions, it is possible to define exponentiation between these special expressions. I have to skip the details, as they are very technical. In particular, that leads to the famous "Euler's identity",

$$\color{green}{e^{i\pi}=-1}.$$

From the latter, we can draw

$$e^{i\pi/2}=i,$$

then

$$\log(i)=i\frac\pi2$$

and finally $$i^i=e^{i\log(i)}=e^{i\cdot i\pi/2}=e^{-\pi/2}.$$

Final remark: those special expressions are known as complex numbers.

The numerous ones who are questioning the "existence" of the complex numbers are on a wrong track. A complex number is just a handy notation for a pair of reals, using the above computation rules. So if you don't question the existence of reals, you have no reason to question that of the complex.

Equivalently, $$(0,1)^{(0,1)}=\left(e^{-\pi/2},0\right).$$

J. W. Tanner
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    To be precise, $i^i$ is multi-valued quantity and $e^{-\pi/2}$ is just a single value among infinitely many possible values. – Bumblebee Mar 16 '21 at 20:25
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    @Bumblebee: I avoided that additional technicality on purpose. –  Mar 16 '21 at 20:26
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    $$ i^i = e^{-n\frac{\pi}{2} }$$ where $n \in \text{odd} \mathbb{Z}$ – Aderinsola Joshua Mar 16 '21 at 20:53
  • It is not more technical than choosing a solution of the equation $x^2=1.$ You can easily modify your answer to obtain all possible values by observing $e^{(2n+1)\pi i}=-1$ for all $n\in\mathbb{Z}.$ – Bumblebee Mar 16 '21 at 20:54
  • @Bumblebee: I don't want to. –  Mar 16 '21 at 21:10