# What is the motivation for complex conjugation?

I have been dealing with complex numbers for few years now. But when I've tried to think about the motivation behind complex conjugation, I was not sure. Let me write what I am working with.

For a complex number $z \in\mathbb{C}$, where $z=\operatorname{Re}z+i\cdot \operatorname{Im}z$, we define complex conjugate of $z$ as $$\overline{z} = \operatorname{Re}z-i\cdot \operatorname{Im}z.$$ Looking at complex numbers in the Gauss plane, this operation is symmetrical around the $x$-axis.

Question Is there any general motivation why we do that? (And after reading the rest of the question, is the motivation I've provided the right one, or are there others?)

I have studied linear algebra, so I know about involution, and adjoints/self-adjoints, where complex conjugation is a very nice example. My guess is that this comes from the fact about the roots of polynomials, where in the quadratic case, we have

$$ax^2 + bx + c = 0$$ and the solutions $$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$ And when $b^2-4ac < 0$, then $\sqrt{b^2-4ac}$ becomes imaginary \begin{align} \sqrt{(-1)\vert b^2-4ac\vert}=\sqrt{(-1)}\sqrt{\vert b^2-4ac\vert}=i\sqrt{\vert b^2-4ac\vert} \end{align} And we get the solutions $$x_{1,2} = \frac{-b}{2a}\pm i\frac{\sqrt{\vert b^2-4ac\vert}}{2a}$$ which only differ in the sign before the imaginary part. Also in the general case, whenever $z$ is the root of $p$, then $\overline{z}$ is also root of $p$. Therefore creating the operation $\overline{\hphantom{a}\cdot\hphantom{a}}$ is justified.

• maths.kisogo.com/index.php?title=Conjugation - this may help. May 19, 2015 at 16:26
• Yes, I would say that the fact that $-i$ is also a root of $x^2+1$ is one of the best motivations. So the fact that complex conjugation exists captures the fact that our choice of $i$ over $-i$ is an arbitrary that doesn't really affect the algebraic structure. May 19, 2015 at 16:29
• Perhaps a more succinct presentation of the same idea: When $i$ is introduced as an imaginary root of $-1$, we have no way to actually discriminate between how $i$ and $-i$ behave in this regard. So we expect (and it can be justified) that switching $-i$ for $i$ throughout complex arithmetic will preserve operations (i.e. be an automorphism of the complex field). May 19, 2015 at 16:31
• If z=r cis(theta), then z bar is r cis(-theta). Conjugation changes the sign of the angle May 19, 2015 at 20:37

One motivation, if you can call it that, is that $i^2=-1$ does not define $i$, because $-i$ also satisfies that equation.

So, there are two elements that could be $i$ and there is no algebraic reason for choosing one over the other. In other words, $\pm i$ are interchangeable, hence conjugation.

Technically, interchangeable means that there is an $\mathbb R$-automorphism of $\mathbb C$ interchanging $i$ and $-i$.

• And furthermore this in the only non trivial $\mathbb{R}$-automorphism we have. Maps which behave analogously to the action of the complex conjugation here occur constantly in Galois theory. May 19, 2015 at 16:34
• Thanks for your answer! I know about automorphisms in general, but what is meant by $\mathbb{R}$-automorphism of $\mathbb{C}$? Does it mean, that $\pi:\mathbb{C} \to \mathbb{C}$ is $\mathbb{R}$-automorphism if $\left.\pi\right|_\mathbb{R} = \operatorname{id}_\mathbb{R}$? May 20, 2015 at 10:19
• @quapka, yes, you have the right definition of $\mathbb R$-automorphism: a ring automorphism that fixes $\mathbb R$ pointwise.
– lhf
May 20, 2015 at 10:40

If $f(x)$ is a polynomial with real coefficients, and $z \in \mathbb C$ is a root of $f$, then $\overline{z}$ is also a root of $f$; in other words complex conjugation acts on the roots of $f$, and we can separate the roots of $f$ into orbits according to this action. An orbit is either a root with $z = \overline{z}$, i.e. a real root, or a pair $\{z, \overline{z}\}$ consisting of a non-real complex number and its complex conjugate. If $z_1, \dots, z_k$ are the real roots and $\{w_1, \overline{w_1}\}, \dots, \{w_r, \overline{w_r}\}$ are the pairs of complex-conjugate roots of $f$, it follows that $f$ factors as

$$f(x) = c \big((x-z_1)(x-z_2)\cdots(x-z_k)\big) \times \big((x^2-2\Re w_1 + |w_1|^2\big)\cdots(x^2-2\Re w_r + |w_r|^2\big))$$

All of the polynomials have real coefficients.

So we see that every polynomial with real coefficients factors as a product of linear factors and quadratic factors, all over the real numbers. All of this thanks to the existence of complex conjugation.

The reason is also to acquire inverses and do division $$\frac{z}{q}=\frac{z\bar{q}}{q\bar{q}}=\frac{z\bar{q}}{|q|^2}$$ Where you get $(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$

• $Im(|q|^2) = 0$ It is a real number. I find it easier to separate Im and Re of $\frac{z\bar{q}}{|q|^2}$ compared to $\frac{z}{q}$, if $Im(q) \neq 0$
– null
May 19, 2015 at 18:05
• Correct, that's why it is so increadibly useful, the name itself comes from the evident relation to normal conjugates May 19, 2015 at 18:06
• I would say that the more fundamental fact here is $x\bar{x} = |x|^2$. A nice formula for the inverse is one of the many nice consequences of this fact. May 19, 2015 at 18:27

After we define addition and multiplication, it is usual to define the division of two complex numbers. Suppose, $u=m+i⋅n$ and $v=p+i⋅q$. And we have already defined multiplication between two complex numbers. So, to solve this issue one way is to make this division in a multiplication. From experience of dealing with surds, here we can try to make the denominator a real number and to maintain equality we need to multiply it in the numerator also. $$uv=\frac{(m+i⋅n)}{(p+i⋅q)}=\frac{(m+i⋅n)⋅(p−i⋅q)}{(p+i⋅q)(p−i⋅q)}$$ Here and later you can see how much important(used frequently) it is to define the number $(p−i⋅q)$ which corresponds to $v=(p+i⋅q)$ as, $\bar{v}$.

I think this will be satisfactory for you. Please feel free to ask for more clarification.