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I have heard "Every equation involving complex numbers and expressions retains its truth value if every complex number/variable is replaced by its complex conjugate."

But, I don't understand how this can be the case. An intuitive reason I recently checked up on is that "the labels $\mathbf{i}$ and $\mathbf{-i}$ are really arbitrary, there is no way to know which is which."

I am not sure I follow. Just because

  1. $-(-i) = i$
  2. $i, -i$ are roots of the same equation $x^{2} + 1 = 0.$

doesn't prove that $\mathbf{i}$ and $\mathbf{-i}$ are truly arbitrary, right?

So, why are the labels $\mathbf{i}$ and $\mathbf{-i}$ arbitrary? Is there any intuitive way of understanding this?

Truth-seek
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  • Just because $,z = 0 ;\iff; \bar z = 0,$. – dxiv Feb 11 '18 at 07:43
  • @dxiv why is that relevant? – Truth-seek Feb 11 '18 at 08:05
  • Because that's what your question essentially asks. Let $,z=x + i y,$ with $,x, y \in \mathbb{R},$, then $\bar z = x - i y$, and so $,z=0 \iff x+iy=0 \iff x=y=0 \iff x- iy = 0 \iff \bar z = 0,$. Replace $,i \mapsto -i,$ and the whole chain of equivalences stays the same, only read backwards. – dxiv Feb 11 '18 at 08:10
  • Yeah. But so, which to call i and which to call -i is arbitrary? – Truth-seek Feb 11 '18 at 08:16
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    I would say this is akin to saying the labels "left" and "right" are arbitrary. Every thing you can say about a plane and whether things are left or right can be switched simply by flipping the plane plane upside down so you are looking at it from the back rather than the front. So just rotate to complex plain around the real axis so that $i$ points down instead of up and everything behaves exactly the same. In fact negative and positive and $<$ and $>$ are equally arbitrary. It's not a very profound or deep statement. – fleablood Feb 11 '18 at 08:19
  • @Truth-seek In the end, it doesn't matter. See here for example – dxiv Feb 11 '18 at 08:21
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    "which to call i and which to call -i is arbitrary? " how could it not be arbitrary? What intrinsic relationship is there between $i$ and the reals so that we can say "that one is the actual square root of negative one; the other is the negative of the square root of negative one"? – fleablood Feb 11 '18 at 08:22

3 Answers3

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The difference between the labels $i$ and $-i$ is whether or not you like to draw your $y$-Axis from bottom to top or from top to bottom. It is the difference whether or not you say clockwise or counter-clockwise the mathematical positive direction of rotation. A matter of convention.

Hyperplane
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I think the statement carries less meaning that a first reading suggests.

I would interpret this is the following way: You are given some variables $z_1,...,z_n$ and an equation. Let $A$ be the set of values of $(z_1,...,z_n)$ such that the equation is true.

Then $(z_1,...,z_n) \in A$ iff $(\overline{z_1},...,\overline{z_n}) \in \overline{A}$.

copper.hat
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Depending on the algebraic structure under consideration, we have units which are elements $u_i$ such that in the same structure is an element $u_i^*$ such that $u_i\cdot u_i^* = 1$. As our algebraic structure gets more involved we may make other demands like the norm of $u_i$ must be $1$. In a sense not unlike how we can factorize integers uniquely into primes up to ordering we can say things like the GCD of 24 and 14 is 2 up to multiplication by a unit since the set of all multiples of $-2$ is the same as the set of all multiples of $2$ (in, say, $\mathbb{Z}$). This doesn't mean all units are indistinguishable. $i$ is different from $-i$ without question.

law-of-fives
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