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.
QuestionIs 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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$\begingroup$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 $-i$.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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 $(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$
$\endgroup$ 3 $\begingroup$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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