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Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

At zero, $[f(h)-f(0)]/h\to 1$, so the complex derivative exists. At a nonzero point, write $r=|z|$. The Cauchy-Riemann equations for $x/(1+r)$ and $y/(1+r)$ force $x^2=y^2$ and $xy=0$, impossible when $r>0$. Hence the function is complex differentiable only at zero and is holomorphic on no nonempty open subset. A complex derivative at one point gives no neighborhood of complex differentiability.

If $f$ is holomorphic on an open set $U$, its conjugate reflection $f^{\ast }$ is holomorphic on $U^{\ast }=\{z:\overline{z}\in U\}$, with $(f^{\ast }{)}^{\prime }(z)=\overline{f^{\prime }(\overline{z})}$. The identity follows by conjugating the difference quotient with increment $\overline{h}$; the two conjugations restore complex differentiability.

If $U$ is invariant under complex conjugation and is connected, and $f$ is real at a set of real points accumulating inside $U$, then $f=f^{\ast }$ by the identity theorem. In particular, an entire function real at every $1/n$ has this symmetry. This is a symmetry statement on an already defined domain; Schwarz reflection also constructs extensions across a boundary.

A holomorphic function on the open set $\Omega$ is complex differentiable at every point of $\Omega$: the limitexists independently of the direction of approach. Write $f(x+iy)=u(x,y)+iv(x,y)$. The Cauchy-Riemann equations areTo prove necessity, approach first with real $h$, obtaining $f^{\prime }(z)=u_x+i{v}_x$. Approach instead with $h=it$ for real $t$, obtainingEquality of real and imaginary parts yields the displayed Cauchy-Riemann equations.

A sufficient condition is that $u$ and $v$ have continuous first partial derivatives throughout $\Omega$ and satisfy the Cauchy-Riemann equations there. Indeed, real differentiability gives, for $h=a+ib$,The Cauchy-Riemann equations imply $f_y=i{f}_x$, so this is $f_{x}h+o(|h|)$. Dividing by $h$ proves complex differentiability and $f^{\prime }(z)=f_x(z)$ at every point. Merely having first partial derivatives and the equations at a point, without a suitable differentiability hypothesis, would not justify this argument.

Write $r=\sqrt{x^2+y^2}>0$. For $u=x/(1+r)$, $v=y/(1+r)$,The two Cauchy-Riemann equations would require $x^2=y^2$ and $xy=0$ simultaneously, impossible when $r>0$. Thus $f$ is not complex differentiable at any nonzero point.

At zero, however,So $f$ is complex differentiable at zero, with derivative one, but is not holomorphic on $\mathbb{C}$ or on any nonempty open subset. Complex differentiability at a single point is weaker than holomorphy in a neighborhood.

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

A conformal map preserves oriented angles between regular smooth curves through each point. For a holomorphic map with nonzero derivative, the derivative also describes the local scale and rotation.

Indeed, if two curves ${\gamma }_1,{\gamma }_2$ meet at $z$ and have nonzero tangent vectors $v_1,v_2$, the real chain rule and complex differentiability giveBecause $f^{\prime }(z)\ne 0$, the image tangents are nonzero andTaking arguments proves preservation of oriented angles. Multiplication by $f^{\prime }(z)$ rotates every tangent by $\arg f^{\prime }(z)$ and scales every length by $|f^{\prime }(z)|$. Thus the hypothesis gives conformality at every point of $U$; global injectivity is not needed for this local conclusion.

At a point $z_0=x+iy$, complex differentiability means thatexists and is independent of the direction of approach. Write $f=u+iv$. Restricting to real $h$ gives $f^{\prime }=u_x+i{v}_x$. Restricting to $h=it$, with real $t\to 0$, givesThese directional limits establish existence of the displayed partial derivatives. Equating their real and imaginary parts gives the Cauchy-Riemann equationsThe argument applies at every point of the given open domain; it does not assume continuity of the partial derivatives beforehand.

For real-direction increments, complex differentiability gives $f^{\prime }=u_x+i{v}_x$. For imaginary-direction increments it gives $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating these expressions gives the Cauchy-Riemann equationsA holomorphic function is smooth, so mixed partial derivatives commute. Differentiating these equations givesThus both real and imaginary parts are harmonic functions:

For an analytic function, the complex differentiability limit is independent of the direction of approach. A real increment gives $f^{\prime }(z)=u_x+i{v}_x$, whereas an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if $u,v$ have continuous first partial derivatives near a point and obey these equations, their real differentiability gives, for a complex increment $h$,Dividing by $h$ proves the complex differentiability there. Continuous first partial derivatives are a sufficient regularity hypothesis for this converse; agreement along the two coordinate directions alone is only a necessary condition.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong real increments $w$ the first factor $\overline{w}/w$ is $1$, while along purely imaginary increments it is $-1$. Thus complex differentiability of $\overline{h}$ requires $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the complex derivative exists and equals zero.

For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. ConsequentlyThese points form a discrete subset of the complex plane and contain no nonempty open set. Every neighbourhood of each one contains points where $F$ is not complex differentiable. Therefore $F$ is holomorphic on no nonempty open set, even though it has complex derivatives at the specified points.

At every $z=x+iy\in U$, the Cauchy-Riemann equations areTo derive them from complex differentiability, approach $z$ along real increments $h$. The complex derivative givesApproaching along imaginary increments $ih$ instead givesThese limits exist because $f$ is holomorphic. Equating their real and imaginary parts proves the equations, together with $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.

At $z_0$, complex differentiability givesWriting $f^{\prime }(z_0)=\lambda e^{i\alpha }$, the assumption $f^{\prime }(z_0)\ne 0$ gives $\lambda >0$. Its real derivative is therefore the similarity $h\mapsto \lambda e^{i\alpha }h$, a rotation through $\alpha$ followed by scaling by $\lambda$.

More explicitly, for two differentiable curves ${\gamma }_1,{\gamma }_2$ with ${\gamma }_j(0)=z_0$ and nonzero tangent vectors, the chain rule givesTaking the complex argument of this ratio shows that the oriented angle is preserved, while nonvanishing ensures both image tangents remain regular. Thus

Complex differentiability gives $u_x=v_y$, $u_y=-v_x$. Differentiating once more, for continuously differentiable first derivatives with the needed second derivatives, yields $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$. Thus both components satisfy Laplace's equation. The result is local on the domain of holomorphy; it makes no assertion across a branch cut.