# Why is there only a complex conjugate, but no real conjugate?

In mathematics one often uses the complex conjugate $$\Bbb C\to\Bbb C,\quad z=a+b\cdot\mathrm{i}\;\;\mapsto\;\; \bar z=a-b\cdot\mathrm{i}$$ This is often described as a a reflection along the real axis. But in analogy one could also define a real conjugate $$\Bbb C\to\Bbb C,\quad z=a+b\cdot\mathrm{i}\;\;\mapsto\;\; \tilde z=-a+b\cdot\mathrm{i}$$ This would be a reflection along the imaginary axis.

However real conjugation is never used. Why is it that complex conjugation is so useful, but real conjugation is not?

• Well, the "real conjugate" is just the negative of the normal conjugate, $z\mapsto-\bar z$, so there's no lack of notation for it. – Akiva Weinberger Jan 12 '16 at 17:08
• Note that the product of a complex number and its complex conjugate is the exact opposite of the product of a complex number and its real conjugate. In that way it's kind of superfluous to have a real conjugate. – Bob Krueger Jan 12 '16 at 17:09
• Note that the complex conjugate satisfies $\overline w\;\overline z=\overline{wz}$, but the real conjugate does not. Also, for any polynomial $p$ with real coefficients, we have $\overline{p(z)}=p(\overline z)$, but this is not true for the real conjugate. (Thus, roots of real polynomials come in complex conjugate pairs.) – Akiva Weinberger Jan 12 '16 at 17:14
• Do you mean imaginary axis? – null Jan 12 '16 at 18:22
• @null Yes. I corrected the text. – asmaier Jan 12 '16 at 18:32

One definition of conjugate arises from the factoring of $a^2 - b^2$ into $(a + b)(a - b)$. But that does not answer the question of why we can obtain the complex conjugate of a complex number only by negating the real part and never by negating the imaginary part. (For that matter, it does not explain why we separate the number into real and imaginary parts in order to obtain a conjugate in the first place.)

But there is another, somewhat different notion of conjugate. Quoting one writer:

Two elements $\alpha, \beta$ of a field $K$, which is an extension field of a field $F$, are called conjugate (over $F$) if they are both algebraic over $F$ and have the same minimal polynomial.

(Barile, Margherita. "Conjugate Elements." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/ConjugateElements.html)

If we take $K$ to be the complex numbers, and $F$ to be the real numbers, then we can verify that $a + bi$ and $a - bi$ ($a, b$ both real) are the two roots of a certain polynomial over $z$, specifically, the solutions for $z$ in the equation

$$z^2 - 2az + a^2 + b^2 = 0,$$

in which the left-hand side is a polynomial over the real numbers (that is, over $F$). That is, $a + bi$ and $a - bi$ fit perfectly the definition of conjugate elements of $\mathbb C$, viewed as an extension field of $\mathbb R$.

What about $a + bi$ and $-a + bi$? These are solutions of the equation $(z - a - bi)(z + a - bi) = 0$; multiplying out the left-hand side, we see that $a + bi$ and $-a + bi$ are solutions for $z$ in

$$z^2 - (2bi)z - a^2 - b^2 = 0.$$

The coefficients of the polynomial on the left-hand side are not all real numbers unless $b = 0$, so it seems that $-a + bi$ cannot be a conjugate of $a + bi$ in the same interesting way that $a - bi$ can.

Historically, complex numbers arose in the process of trying to solve polynomials with real-valued coefficients. Eventually, people decided that complex numbers were actually acceptable roots of such a polynomial. When you have a polynomial with integer coefficients, it always has a factorization into polynomials of the form $(ax + b)$ or $(ax^2 + bx + c)$, where all the coefficients $a, b, c$ are real numbers. Of course $(ax + b)$ has just one real root (and no other roots), but the roots of $(ax^2 + bx + c)$ are precisely a pair of complex conjugates.

The complex conjugate is an automorphism that maps $\mathbb C \to \mathbb C$. There are no non-trivial automorphisms $\mathbb R \to \mathbb R$.

Edit: $\mathbb C$ can be defined as the splitting field (https://en.wikipedia.org/wiki/Splitting_field) $\mathbb R(X)/(X^2+1)$. The roots of $X^2+1=0$ are symmetric, so $X=i$ or $X=-i$ is an arbitrary choice.

Edit2: an automorphism $\alpha$ observes addition and multiplication, such that for every $a,b$ it is true that $\alpha(a+b)=\alpha(a)+\alpha(b)$ and $\alpha(a\cdot b)=\alpha(a)\cdot\alpha(b)$. This does not hold for the OP's proposed "real conjugate".

Edit3: for the proposed conjugate ($\Bbb C\to\Bbb C,\quad z=a+b\cdot\mathrm{i}\;\;\mapsto\;\; \tilde z=-a+b\cdot\mathrm{i}$), this would result into $i=\tilde{i}=\widetilde{1\cdot i}=\tilde{1}\cdot\tilde{i}=-1\cdot i=-i$ when assuming it would work as an automorphism (or at least as an homomorphism). This gives a contradiction.

• Aren't the roots of $X^2 -1 = 0$ also symmetric, so $X = 1$ and $X=-1$ is an arbitrary choice? – asmaier Jan 12 '16 at 18:36
• @asmaier no, because 1 is the multiplicative identity, -1 is not. So exchanging the two gives different results. And "symmetric" is perhaps not the best word for it, will try to think of better wording. – Gyro Gearloose Jan 12 '16 at 18:38
• Are you sure this answers the OP's question? Their $\sim$ operator is not $\mathbb{R} \to \mathbb{R}$, but $\mathbb{C} \to \mathbb{C}$; it's just the usual conjugate, but "flipped" (so to speak). – Brian Tung Jan 12 '16 at 18:38
• @BrianTung, not 100%, but will find out. – Gyro Gearloose Jan 12 '16 at 18:39
• @BrianTung 99% sure now, but still awaiting @_asmeiers comment, but calculating the time that anyone's brain needs to suck in any given answer is probably NP-hard to calculate, and the associated theory, depending on how far one will dip in, can consume a real lot of time. – Gyro Gearloose Jan 12 '16 at 21:26