Editing Algebraic function (section)

=== The role of complex numbers ===
From an algebraic perspective, complex numbers enter quite naturally into the study of algebraic functions.  First of all, by the [[fundamental theorem of algebra]], the complex numbers are an [[algebraically closed field]].  Hence any [[polynomial]] relation ''p''(''y'',&thinsp;''x'') = 0  is guaranteed to have at least one solution (and in general a number of solutions not exceeding the degree of ''p'' in ''y'') for ''y'' at each point ''x'', provided we allow ''y'' to assume complex as well as [[Real number|real]] values.  Thus, problems to do with the [[domain of a function|domain]] of an algebraic function can safely be minimized.

[[Image:y^3-xy+1=0.png|thumb|A graph of three branches of the algebraic function ''y'', where ''y''<sup>3</sup>&nbsp;&minus;&nbsp;''xy''&nbsp;+&nbsp;1&nbsp;=&nbsp;0, over the domain 3/2<sup>2/3</sup> < ''x'' < 50.]]
Furthermore, even if one is ultimately interested in real algebraic functions, there may be no means to express the function in terms of addition, multiplication, division and taking ''nth'' roots without resorting to complex numbers (see [[casus irreducibilis]]).  For example, consider the algebraic function determined by the equation

:<math>y^3-xy+1=0.\,</math>

Using the [[cubic formula]], we get

:<math>
y=-\frac{2x}{\sqrt[3]{-108+12\sqrt{81-12x^3}}}+\frac{\sqrt[3]{-108+12\sqrt{81-12x^3}}}{6}.
</math>

For <math>x\le \frac{3}{\sqrt[3]{4}},</math> the square root is real and the cubic root is thus well defined, providing the unique real root. On the other hand, for <math>x>\frac{3}{\sqrt[3]{4}},</math> the square root is not real, and one has to choose, for the square root, either non-real square root. Thus the cubic root has to be chosen among three non-real numbers. If the same choices are done in the two terms of the formula, the three choices for the cubic root provide the three branches shown, in the accompanying image.

It may be proven that there is no way to express this function in terms of ''nth'' roots using real numbers only, even though the resulting function is real-valued on the domain of the graph shown.

On a more significant theoretical level, using complex numbers allows one to use the powerful techniques of [[complex analysis]] to discuss algebraic functions.  In particular, the [[argument principle]] can be used to show that any algebraic function is in fact an [[analytic function]], at least in the multiple-valued sense.

Formally, let ''p''(''x'',&thinsp;''y'') be a complex polynomial in the complex variables ''x'' and ''y''.  Suppose that
''x''<sub>0</sub>&thinsp;∈&thinsp;'''C''' is such that the polynomial ''p''(''x''<sub>0</sub>,&thinsp;''y'') of ''y'' has ''n'' distinct zeros.  We shall show that the algebraic function is analytic in a [[Neighborhood (mathematics)|neighborhood]] of ''x''<sub>0</sub>.  Choose a system of ''n'' non-overlapping discs Δ<sub>''i''</sub> containing each of these zeros.  Then by the argument principle

:<math>\frac{1}{2\pi i}\oint_{\partial\Delta_i} \frac{p_y(x_0,y)}{p(x_0,y)}\,dy = 1.</math>

By continuity, this also holds for all ''x'' in a neighborhood of ''x''<sub>0</sub>.  In particular, ''p''(''x'',&thinsp;''y'') has only one root in Δ<sub>''i''</sub>, given by the [[residue theorem]]:

:<math>f_i(x) = \frac{1}{2\pi i}\oint_{\partial\Delta_i} y\frac{p_y(x,y)}{p(x,y)}\,dy</math>

which is an analytic function.