Editing Laplace's equation (section)

===Analytic functions===
<!-- This section is linked from [[Complex analysis]] -->
The real and imaginary parts of a complex [[analytic function]] both satisfy the Laplace equation. That is, if {{math|1=''z'' = ''x'' + ''iy''}}, and if 
<math display="block">f(z) = u(x,y) + iv(x,y),</math>
then the necessary condition that {{math|''f''(''z'')}} be analytic is that {{math|''u''}} and {{mvar|''v''}} be differentiable and that the [[Cauchy–Riemann equations]] be satisfied:
<math display="block">u_x = v_y, \quad v_x = -u_y.</math>
where {{math|''u<sub>x</sub>''}} is the first partial derivative of {{math|''u''}} with respect to {{mvar|x}}.
It follows that 
<math display="block">u_{yy} = (-v_x)_y = -(v_y)_x = -(u_x)_x.</math>
Therefore {{math|''u''}} satisfies the Laplace equation. A similar calculation shows that {{math|''v''}} also satisfies the Laplace equation. 
Conversely, given a harmonic function, it is the real part of an analytic function, {{math|''f''(''z'')}} (at least locally). If a trial form is
<math display="block">f(z) = \varphi(x,y) + i \psi(x,y),</math>
then the Cauchy–Riemann equations will be satisfied if we set
<math display="block">\psi_x = -\varphi_y, \quad \psi_y = \varphi_x.</math>
This relation does not determine {{math|''ψ''}}, but only its increments:
<math display="block">d \psi = -\varphi_y\, dx + \varphi_x\, dy.</math>
The Laplace equation for {{math|''φ''}} implies that the integrability condition for {{math|''ψ''}} is satisfied:
<math display="block">\psi_{xy} = \psi_{yx},</math>
and thus {{math|''ψ''}} may be defined by a line integral. The integrability condition and [[Stokes' theorem]] implies that the value of the line integral connecting two points is independent of the path. The resulting pair of solutions of the Laplace equation are called '''conjugate harmonic functions'''. This construction is only valid locally, or provided that the path does not loop around a singularity. For example, if {{mvar|r}} and {{mvar|θ}} are polar coordinates and
<math display="block">\varphi = \log r,</math>
then a corresponding analytic function is
<math display="block">f(z) = \log z = \log r + i\theta.</math>

However, the angle {{mvar|θ}} is single-valued only in a region that does not enclose the origin.

The close connection between the Laplace equation and analytic functions implies that any solution of the Laplace equation has derivatives of all orders, and can be expanded in a [[power series]], at least inside a circle that does not enclose a singularity. This is in sharp contrast to solutions of the [[wave equation]], which generally have less regularity{{citation needed|date=July 2020}}.

There is an intimate connection between power series and [[Fourier series]]. If we expand a function {{math|''f''}} in a power series inside a circle of radius {{mvar|R}}, this means that
<math display="block">f(z) = \sum_{n=0}^\infty c_n z^n,</math>
with suitably defined coefficients whose real and imaginary parts are given by 
<math display="block">c_n = a_n + i b_n.</math>
Therefore
<math display="block">f(z) = \sum_{n=0}^\infty \left[ a_n r^n \cos n \theta - b_n r^n \sin n \theta\right] + i \sum_{n=1}^\infty \left[ a_n r^n \sin n\theta + b_n r^n \cos n \theta\right],</math>
which is a Fourier series for {{math|''f''}}.  These trigonometric functions can themselves be expanded, using [[De Moivre's formula#Formulas for cosine and sine individually|multiple angle formulae]].