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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.

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  • $\begingroup$ maths.kisogo.com/index.php?title=Conjugation - this may help. $\endgroup$ – Alec Teal May 19 '15 at 16:26
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    $\begingroup$ 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. $\endgroup$ – Hayden May 19 '15 at 16:29
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    $\begingroup$ 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). $\endgroup$ – hardmath May 19 '15 at 16:31
  • $\begingroup$ If z=r cis(theta), then z bar is r cis(-theta). Conjugation changes the sign of the angle $\endgroup$ – Faraz Masroor May 19 '15 at 20:37
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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 $-i$.

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    $\begingroup$ 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. $\endgroup$ – b00n heT May 19 '15 at 16:34
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    $\begingroup$ 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 $\left.\pi\right|_\mathbb{R} = \operatorname{id}_\mathbb{R}$? $\endgroup$ – quapka May 20 '15 at 10:19
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    $\begingroup$ @quapka, yes, you have the right definition of $\mathbb R$-automorphism: a ring automorphism that fixes $\mathbb R$ pointwise. $\endgroup$ – lhf May 20 '15 at 10:40
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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.

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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 $(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$

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  • $\begingroup$ $Im(|q|^2) = 0$ 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 $Im(q) \neq 0$ $\endgroup$ – null May 19 '15 at 18:05
  • $\begingroup$ Correct, that's why it is so increadibly useful, the name itself comes from the evident relation to normal conjugates $\endgroup$ – Zelos Malum May 19 '15 at 18:06
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    $\begingroup$ 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. $\endgroup$ – Sasho Nikolov May 19 '15 at 18:27
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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.

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