Polynomials have a canonical form $\sum a_n x^n$ which makes it easy to understand what a polynomial is. Yet rational expressions can often wear many disguises, where it's not obvious that two rational expressions are the same, or that a complex rational expression is really something recognizable.
Is there a canonical form to put rational expressions into? If not, are there methods which can be used to understand or identify what a rational expression really is?
A few examples:
Look at What do the fixed points and symmetry of $f(x) = \frac {ax +b} {cx + d}$ tell us? Geometry, symmetry, and limits. . It was not initially obvious, at least to me, that $$\frac{ax+b}{cx+d}$$ equals $$S \cdot \frac 1 {x - V_a} + H_a$$ for some constants $S, V_a, H_a$. It wasn't even obvious to search in that direction.
Or the comments at Prove $\frac {uv} {(u+v)^2}$ is real and $\frac {u+v}{u-v}$ imaginary when $|u| = |v| = 1$ . Most involve converting the rational expression into something else.
Or even the (almost obvious) equality $$\frac{x}{1 - \frac{x-1}{x}} = x^2.$$Yes, it's obvious once you see it, or even once you say "How do I simplify this?". But I wouldn't see $\frac{x}{1 - \frac{x-1}{x}}$ and immediately notice that it's really just $x^2$. How do I notice that?
One canonical form is simply $p(x)/q(x)$, where $p(x)$ and $q(x)$ are polynomials. To be in simplest form, they ought to share no (nonconstant) factors, otherwise we could cancel the factors. If the coefficients of the polynomials are rational, we can even multiply through to ensure the coefficients are integers. Typically the leading terms will both be positive, and if necessary a minus sign will be put out in front of the fraction rather than putting it in one of the numerator or denominator.
A "complex rational expression" is a rational expression which contains (possibly other complex) rational expressions, for example $x/(1-(x-1)/x)$. These can be simplified to plain-old rational expressions so long as you know how to add, subtract, multiply, and divide fractions. So in that example, you can simplify the denominator $1-(x-1)/x$ to a single fraction, and then divide $x$ by the result. Another little trick in that particular case is to note $(x-1)/x=1-1/x$: the numerator and denominator differ by a constant, so distributing the division allows us to compare its size with the ratio $1$ itself, which is useful because we are then going to be subtracting it from $1$. When $x=10$ this denominator is $1-0.9=0.1$, for example. We can simplify $x/(1/x)$ by multiplying-by-the-reciprocal or multiplying the top and bottom by $x$. (Think about the ratio is what-you-need-to-multiply-the-bottom-by-to-get-the-top: we multiply $1/x$ by $x$ to get $1$, then multiply $1$ by $x$ to get $x$, so we multiply $1/x$ by $x^2$ to get $x$.)
Anything you can do to simplify a fraction, you can also do in reverse if the situation calls for it. (Sometimes just experimenting leads places.) Some limit definitions of derivatives we encounter in beginning calculus can be done by "rationalizing the numerator" (instead of the denominator), for example. Another example is $uv/(u+v)^2$: if we divide the top and the bottom by $uv$ for example, we get $1/((u^{-1}+v^{-1})(u+v))$. You can recognize $(u+v)/(uv)=u^{-1}+v^{-1}$ if you've ever added reciprocals before, and doing this is useful because $u^{-1}=\bar{u}$, $v^{-1}=\bar{v}$, allowing us to write the denominator as $(\overline{u+v})(u+v)=|u+v|^2$. (As for showing $(u-v)/(u+v)$ is imaginary algebraically, the best you can do is probably the algebra in the linked question. The geometric explanation in the answer there is better though.)
Another form that's very useful in complex analysis (using the residue theorem to calculate contour integrals, for example) is the partial fraction decomposition. Over a chosen field (i.e. number system like $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$) there are two possible representations:
You can actually do this with rational numbers too! For example:
$$ \begin{array}{ccl} \displaystyle \frac{1115}{168} & = & \displaystyle 5+\frac{7}{2^3}+\frac{1}{3}+\frac{3}{7} \\ & = & \displaystyle 5 + \Bigl(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\Bigr)+\frac{1}{3}+\frac{3}{7} \end{array} $$
Observe $168=2^3\cdot3\cdot7$ and the numerators are less than the denominators. For rational numbers we find these values using modular arithmetic. For polynomials over $\mathbb{C}$, the only irreducible polynomials are linear (fundamental theorem of algebra) and we can similarly solve for these values by removing powers $(x-\lambda)^e$ from the denominator and evaluating at $x=\lambda$. (These techniques are actually analogous, an intuition we glean from algebraic geometry and function field stuff.) For example:
$$ \begin{array}{ccl} \displaystyle\frac{x^6-x^4+x^3+5x^2-3x+5}{x^4-2x^2+1} & = & \displaystyle x^2+1+\frac{x+1\,}{(x-1)^2}+\frac{3}{(x+1)^2} \end{array} $$
We can break down $A(x)/\pi(x)^e$ into a sum of $a(x)/\pi(x)^f$s ($f=1,\cdots,e$) by polynomial-long-dividing $A(x)$ by $\pi(x)$ (over and over again as necessary). For instance, dividing $x+1$ by $x-1$ yields $x+1=1(x-1)+2$ (quotient is $1$, remainder is $2$), so $(x+1)/(x-1)^2=1/(x-1)+2/(x-1)^2$. Note polynomial long division is basically the same as the long division we learn in grade school, except we use "base-$x$" (and allow arbitrary "digits," i.e. coefficients) instead of "base-ten" (with digits from $0,\cdots,9$).
In the case of a Mobius transformation $(ax+b)/(cx+d)$, we can use polynomial long division, and then divide top/bottom by the leading coefficient, to write it as $h+s/(x-v)$, which is a composition of homotheties (aka dilations), translation, and inversion (this is useful for understanding the geometry of Mobius transformations acting on the Riemann sphere, or extended complex plane, which contains the real projective line).