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An algebraic function is a function whose values satisfy a nonzero polynomial relation $P(z,f(z))=0$. In complex analysis one usually studies a local holomorphic branch of this relation; the relation itself need not specify one globally single-valued function.

For example, $w^2-z=0$ gives two square root branches away from zero. On a slit domain, one can choose a holomorphic branch, but following it around zero exchanges the two values. This is why Riemann surfaces arise naturally when studying an algebraic function. Every rational function $f=p/q$ is an algebraic function, since $q(z)f(z)-p(z)=0$ wherever $q\ne 0$. In contrast, the exponential function is a transcendental function.

A transcendental function is an analytic function that is not an algebraic function. Explicitly, on a domain $U$, there is no nonzero polynomial $P(z,w)$ with complex coefficients for whichA rational function is algebraic, and a local branch of $\sqrt{z}$ is algebraic because it satisfies $w^2-z=0$. The exponential function is transcendental. This concerns a relation between functions, rather than the arithmetic nature of a single value: $e^0=1$ even though $z\mapsto e^z$ is transcendental.

Here is a direct proof for the exponential function. Suppose ${\sum }_{k=0}^m{p}_k(z)e^{kz}=0$ on $U$, with polynomials $p_k$ and $p_m\ne 0$. The expression is an entire function, so the identity theorem for holomorphic functions extends the identity to all complex $z$. For positive real $x$, divide by $e^{mx}$:By exponential domination of polynomials, the right side tends to zero as $x\to \infty$. A nonzero polynomial cannot tend to zero along the positive real axis, giving a contradiction. The case $m=0$ is already impossible because $p_0$ would vanish identically. Thus no such algebraic relation exists.