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My math professor made what appears to me to be a sweeping remark that if we have a long and complicated equation involving the imaginary unit 'i', and we replace it everywhere with '-i', as factors of terms and even in the exponents,the equation would still remain valid. The reason he provided is that both i and -i are defined in the same unique manner which is that they both follow only from the equation x²=1. Can anyone make it more clearer?

I checked the Wikipedia page also and it was quite useful but I still only dimly understand it. Main points which I don't understand are(quoted from Wikipedia):

  1. "After all, although −i and +i are not quantitatively equivalent (they are negatives of each other), there is no algebraic difference between +i and −i, as both imaginary numbers have equal claim to being the number whose square is −1."

  2. "The distinction between the two roots x of x²+ 1 = 0, with one of them labelled with a minus sign, is purely a notational relic." What exactly does that mean?

  3. "+i and -i are not quantitatively equivalent but are algebraically equivalent." I think that begs the question.

Any help would be greatly appreciated.

Ad Nazir
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  • Point 2: you could have reinvented everything with $j=-i$ and $-j=i$ and you'd be none the wiser, esp. with respect to $x^2+1=0$. – Randall Sep 12 '22 at 18:00
  • It's not necessarily true if you consider polynomials with complex coefficients. For example, $z-i=0$ is not equivalent to $z+i=0$. – Vasili Sep 12 '22 at 18:17
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    complex conjugation is a field automorphism – David Lui Sep 12 '22 at 18:20
  • I think the key concept is that we can't make $i>0$ or $i<0$, expanded thoughts in my answer – Alan Sep 12 '22 at 18:21
  • @Vasili But if you change i to negative i in the second equation the z would also change to make the equation valid, i think – Ad Nazir Sep 12 '22 at 20:20

1 Answers1

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The idea here is that we have strong notions for why $1$ and $-1$ are different in all kinds of ways, we have a natural ordering on the real numbers by $<$ which is compatible with the field operations: $$a<b\implies a+c<b+c$$ and $$a<b \text{ and } c>0 \implies ac<bc$$

This lead us to the natural consequence that any positive number had 2 square roots, because multiplying a negative times itself just turned itself positive, i.e. the solutions to $x^2=4$ are $2$ and $-2$. We still have a lot of strong other things from the ordered field axioms that distinguish positive and negative.

When we want to add numbers to our system so $x^2=-1$ has a solution, it turns out there is no way to make the complex numbers (The smallest algebraicly closed field containing the reals, i.e. every polynomial equation has a solution) into an ordered field. We could order the complex numbers, but not in a way that is compatible with the arithmetic operations.

So, we invent a symbol as our solution, we call it $i$ and toss it in with our real numbers. Problem: Now we no longer have the nice properties such as every number has an additive inverse, so we have to create a $-i$ as well. Equal problems, we no longer have you can multiply any two numbers and get a number, what's $5\cdot i$? Also you can no longer add two numbers and get a number, what's $5+i$? So we can't just toss in $i$ and call it a day. Instead we take the algebraic closure of $\mathbb{R}\cup \{i\}$, in other words, the smallest set of numbers such that we can do all the usual arithmetic operations and have it work. It turns out this is the complex numbers, all numbers of the form $a+bi$, where $a,b\in \mathbb{R}$

Since there is no order though, as far as all of the equations are concerned, if we swapped out what symbol we called $'i'$ with what our system calls $'-i'$, there would be absolutely no change.

After all, we cannot claim $i>0$ or $i<0$ in a meaningful way, which is how we distinguish positive and negative real numbers

I could start with $\mathbb{R} \cup \{-i\}$, and the system would work identically.

Now, once we add a geometric representation of the complex numbers and identify $i$ as the vertical unit, we get a geometric difference between $i$ and $-i$, that being multiplying by $i$ is a counterclockwise 90 degree angle, whereas $-i$ is a clockwise 90 degree angle. But that's just a visual representation and not inherent to the formulae.

Alan
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