# Why do we negate the imaginary part when conjugating?

For $z=x+iy \in \mathbb C$ we all know the definition for the "conjugate" of $z$, $\bar{z}=x-iy$. Geometrically this is the reflection of $z$ across the $y$ axis.

My question is: couldn't we have defined $\underline{z}=-x+iy$ instead? (this is the reflection across the $x$ axis) Is there anything wrong with it?

I agree that the formula $|z|^2=z \bar{z}$ looks better than $|z|^2=-z \underline{z}$, but are there any more serious problems?

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The main problem, for me at least, is that this new operator does not fix the real axis. The nice thing about the complex conjugate is that you can do lots of complex analysis without altering the real axis. The real axis can sit naturally within the complex plane and be compatible. –  Fly by Night Mar 26 '13 at 21:11
No problem, as long as we are willing to put in a minus sign whenever it is needed, which is a lot. If I were in the business of selling minus signs, I would strongly support defining the conjugate as $-x+iy$. –  André Nicolas Mar 26 '13 at 21:12
@AndréNicolas If I were in the business of selling minus signs, I would support $-x-i(-y)$ :) –  rschwieb Mar 26 '13 at 21:18
@rschwieb: You are right, my business instincts were inadequate. –  André Nicolas Mar 26 '13 at 21:28
Shouldn't that be reflection about the $x$-axis then $y$-axis, not the other way round? Although I do note you use the word "across" and not "about". –  pbs Sep 6 '13 at 14:52

No, the nice thing about conjugation is that it is an automorphism: $\overline{zw} = \bar z\bar w$ and $\overline{z+w}=\bar z+\bar w$.

At heart, what conjugacy shows you is that, while $i$ and $-i$ are different complex numbers, they have exactly the same behavior. That's not surprising, because we define $i$ so that $i^2=-1$, but then $(-i)^2=-1$. How do we distinguish these two square roots? What if we defined the complex numbers as $a+bi+cj$ where $i+j=0$ and $i^2=j^2=-1$? Then which would be the "primary" square root of $-1$? There is really no way to tell. With positive real numbers, we can always pick the positive square root, but we can't define "positive" in the complex numbers. This yields a duality in the complex numbers, represented by conjugation.

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Could you expand on "This yields a duality in the complex numbers, represented by conjugation."? –  Pedro Tamaroff Mar 26 '13 at 22:33
$\bar z$ is a dual of $z$ by replacing $i$ with $-i$. Not sure what other expansion you want @PeterTamaroff. I'm not using dual in the linear algebra sense, just a casual language sense. Specifically, that there are two ways of looking at the complex numbers that "look" identical - a duality. –  Thomas Andrews Mar 26 '13 at 22:39
Oh, OK. ${}{}{}{}$ –  Pedro Tamaroff Mar 26 '13 at 22:44
Well of course it's possible to elaborate on this notion of duality: the conjugation operator can be seen as the extension a map which sends roots of the polynomial $X^2 + 1$ to roots of the same, and it is the only such map other than the identity. The study of these maps is the main object of Galois theory. –  cody Mar 28 '13 at 18:58
very nice answer! –  Lior B-S Mar 28 '13 at 22:03

The notion of a conjugate can be seen from an algebraic number theory perspective, and then the "usual" complex conjugate is just a special case of this (so in particular this gives motivation as to why it has the form it does.)

Say we have a number $\alpha$ which is the root of a polynomial with rational coefficients. Then define the minimal polynomial of $\alpha$ over $\mathbf{Q}$ to be the polynomial of smallest degree for which $\alpha$ is a root (let's also say the leading coefficient is $1$ so this polynomial is unique). Once this is defined, we say the conjugates of $\alpha$ are the other solutions to this polynomial.

## Example

Consider $1+\sqrt{2}$. The minimal polynomial of this is $x^2-2x-1$ (plug it in and you get $0$).

You can use the quadratic formula to factor $x^2-2x-1$ (or you can eyeball it if you are a factoring prodigy), and what you get is$$(x-(1+\sqrt{2}))(x-(1-\sqrt{2})),$$

so using our algebraic definition of conjugate, $1+\sqrt{2}$ and $1-\sqrt{2}$ are conjugates. This is typical for the situation with quadratics; just consider the quadratic formula and that $\pm$ which shows up right before the radical. If the roots are complex, we even get$$\frac{-b}{2a}\color{red}{\pm\,i}\cdot\frac{\sqrt{b^2-4ac}}{2a},$$which suggests that the conjugate of $a+bi$ should be $a-bi$.

Something else floating around these ideas is the notion of a norm, which you mentioned in your question. Indeed, if we multiply the "conjugates" $(a+b\sqrt{2})(a-b\sqrt{2})$, you get $a^2-2b^2$, which has no pesky square roots. This is a norm for the element $a+b\sqrt{2}$ in some extension of rationals, and norms are useful because (among other things) they are in the base field (i.e. they are rational).

Similarly, the norm of $z=a+bi$ is as you mentioned $z\cdot \overline{z}$, which turns out to be real. Of course, it's real if you define the conjugate as your alternative in the question, but the moral of the story is that we want the "conjugate" to generalize nicely to these other algebraic situations where we are extending by things like $\sqrt{2}$ instead of $i$.

For complex numbers (and other so-called quadratic extensions) this means we need to flip the sign of the thing we are extending by (so $i$ in this case).

In case no one believes this, see here and references therein.

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The property of a conjugate is that the argument of the conjugate is the additive inverse of the argument of the original. If $z$ is 30 degrees clockwise from the real axis, then $\bar z$ is 30 degrees counter-clockwise. The upshot of this is that when you multiply the numbers together, you get a real number, because the arguments cancel out. (As you likely know, multiplication of complex numbers results in addition of arguments.)

Another nice thing about the conjugate is that it basically goes away when we restrict to the real numbers. The conjugate of a real is just that real. So in some cases, $z\bar z$ is an extension of $z^2$ into the complex numbers. For instance distance from the origin is $\sqrt{z^2}$ if $z$ is real, but we generalize it as $\sqrt{z\bar z}$ in the complex numbers. It "specializes back" to $\sqrt{z^2}$.

Nice, simple things like this break if you try to find some other concept for the complex conjugate.

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Consider the equation $z^2 + 1$. As you probably know, $i$ is a root of this equation. In fact, this is the most commonly used definition of $i$: a number such that $i^2=-1$.

But $-i$ is also a root of $z^2+1$, so why not choose this one? Ok. Let us write $i'=-i$. Since $i$ and $i'$ share the same definition, "everything" that is satisfied by one of them has to be satisfied by the other one. For example, since $(1+i)^4 = -4$, we must have $(1+i')^4 = -4$; this is true!

When I say "everything", I actually mean every polynomial equation. The general principle is the following: if $i$ is a root of some polynomial $P(z)$ with real coefficients, then so $i'=-i$. In the previous example, the polynomial would be $P(z) = (1+z)^4 + 4$.

This principle is more powerful than it looks at first sight. For example, if $a+ib$ (with $a,b \in \Bbb R$) is a root of some polynomial $Q(z)$ with real coefficients, then $i$ is a root of $P(z) = Q(a+zb)$. We deduce that $-i$ is a root of $P(z)$, that is $a-ib$ is a root of $Q(z)$. In the same way, you can see that $$(a_1+ib_1) + (a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1) + (a_2-ib_2)=(a_3-ib_3)$$ with $P(z) = (a_1+zb_1) + (a_2+zb_2)-(a_3+zb_3)$, and $$(a_1+ib_1)(a_2+ib_2)=(a_3+ib_3) \iff (a_1-ib_1)(a_2-ib_2)=(a_3-ib_3)$$ (which polynomial would you take?)

The notion of conjugate is just a way to formalize all these "symmetries" between $i$ and $-i$. For instance, the two last equivalences write $$\overline{z_1}+ \overline{z_2} = \overline{z_1+ z_2} ,\qquad \overline{z_1}\times \overline{z_2} = \overline{z_1\times z_2}$$

Remark: the same kind of things could be done with $\sqrt{2}$ and $-\sqrt{2}$ and polynomial equations with rational coefficients. The general idea behind all of this is is that of Galois Theory.

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If we want complex numbers $z$ and their conjugates $\bar{z}$ to have the property that whenever $z$ is a root of a polynomial $f(t)$ with real coefficients, then so is $\bar{z}$ a root of the same polynomial $f(t)$, then we need to define $\bar{z}$ as $x-iy$. This is because $f(t)$ must be a multiple of $$q(t) = (t - z)(t-\bar{z}) = t^2 - (z+\bar{z})t + |z|^2$$ and the coefficient of $t$ in $q(t)$ is not real if we use your definition of conjugate, but is indeed real if the standard definition $\bar{z} = x-iy$ is used.

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(Hint: it would pay off to think simply: $i$ hmmmm...)