Given \(u\) and \(v\) explicitly
Cauchy–Riemann equations. Write \(f(x+iy) = u(x,y) + iv(x,y)\) and check \(u_x = v_y\) and \(u_y = -v_x\).
If \(u\) and \(v\) have continuous partial derivatives on an open set and satisfy the equations there, then \(f\) is holomorphic on that set. At a single point the equations are weaker than complex differentiability: \(f(x+iy) \coloneqq \sqrt{\abs{xy}}\) satisfies them at \(0\), and along \(y=x\) its difference quotient at \(0\) is \(\abs x/\bigl((1+i)x\bigr)\), which has no limit as \(x\to 0\). In polar coordinates the equations read \(u_r = \frac1r v_\theta\) and \(v_r = -\frac1r u_\theta\).
Equivalently, \(f\) is holomorphic exactly when \(\delbar f = 0\), the Cauchy–Riemann equations written as one equation. See The Cauchy–Riemann equations.
Given a power series
Radius of convergence. A power series is holomorphic on the open disc of convergence, and every holomorphic function is locally given by such a series. See Power series.
Given a limit or an integral
Morera’s theorem. If \(f\) is continuous and \(\int_{\bd T} f = 0\) for every triangle \(T\), then \(f\) is holomorphic. This applies to functions without a formula to differentiate: a locally uniform limit of holomorphic functions, a series of them, or an integral depending holomorphically on a parameter. See Morera and converses.
Given a formula built from holomorphic functions
Sums, products, quotients with nonvanishing denominators, and compositions of holomorphic functions are holomorphic, and \(e^z\), \(\sin z\), \(\cos z\), and polynomials are entire.
Obstructions to holomorphy
(Holomorphic and not)
- \(f(z) \coloneqq \abs z\) is not holomorphic.
- \(f(z) \coloneqq \arg z\) is not holomorphic.
- \(f(z) \coloneqq \Re z\) is not holomorphic.
- \(f(z) \coloneqq \Im z\) is not holomorphic.
- \(f(z) = 1/z\) is holomorphic on \(\CC\smz\), and not on \(\CC\).
- \(f(z) = \bar z\) is real differentiable but not holomorphic: \[ {f(z_0 + h) - f(z_0) \over h } = {\bar h \over h} = {re^{-i\theta} \over re^{i\theta}} = e^{-2i\theta}, \] which depends on the direction \(\theta\) of approach, so the limit as \(h\to 0\) does not exist.
Direction dependence. The difference quotient has different limits along different rays, as for \(\bar z\). A nonconstant real-valued function of \(z\), such as \(\Re z\), \(\Im z\), \(\abs z\), or \(\arg z\), fails in this way.
A branch point. \(\sqrt z\) and \(\log z\) have no continuous branch on any punctured neighborhood of \(0\), so the obstruction is topological rather than pointwise.
(The square root has no holomorphic branch on the circle)
With the branch \((e^{i\theta})^{1/2} = e^{i\theta/2}\) for \(0\le\theta\le 2\pi\), \[ \begin{aligned} \int_{S^1} z^{1/2} \dz &= \int_0^{2\pi} e^{i\theta/2}\, ie^{i\theta} \dtheta \\ &= i \int_0^{2\pi} e^{3i\theta/2}\dtheta \\ &= i \qty{2\over 3i} e^{3i\theta/2}\evalfrom_{0}^{2\pi} \\ &= -{4\over 3}. \end{aligned} \] With the branch \(e^{i\theta/2}\) for \(-\pi\le\theta\le\pi\), \[ \int_{-\pi}^{\pi} e^{i\theta/2}\, ie^{i\theta} \dtheta = {2\over 3}\qty{ e^{3\pi i/2} - e^{-3\pi i/2}} = -{4i\over 3}. \] Each branch is discontinuous at one point of \(S^1\), and the two integrals differ. If \(g\) were a holomorphic branch of \(z^{1/2}\) on a neighborhood of \(S^1\), then \(g' = g/(2z)\), so \(h(z)\coloneqq \frac23 z\,g(z)\) would satisfy \(h' = g\), and the integral of \(g\) over the closed curve \(S^1\) would vanish.
- The domain. \(1/z\) is holomorphic on every open set not containing \(0\), so holomorphy is a statement about a function on a specified open set.
Warnings
\(\sin z\) and \(\cos z\) are unbounded on \(\CC\): they are nonconstant and entire, so this follows from Liouville’s theorem. For \(y\in\RR\), \(\abs{\sin(iy)} = \abs{\sinh y}\).
A map whose real Jacobian is invertible at every point need not be holomorphic: \(f(z) = \bar z\) has Jacobian \(\operatorname{diag}(1,-1)\) and fails the Cauchy–Riemann equations. Conversely, a holomorphic \(f\) with \(f'(z_0)\neq 0\) is conformal at \(z_0\).