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Let $U\subseteq \mathbb{C}$ be open, let $L$ be a straight line, and let $f:U\to \mathbb{C}$ be continuous on $U$ and holomorphic on $U\setminus L$. Then $f$ is holomorphic on all of $U$. Near a point of $L$, work in a disk whose closure lies in $U$. Split a small filled triangle along $L$ and triangulate the two resulting convex polygons. Translate each triangle slightly into its corresponding open half-plane, keeping it in $U$. Its contour integral is zero by Cauchy's integral theorem. On each segment the integral is the endpoint difference times the integral of $f$ along the unit-interval parametrization; uniform continuity on a containing compact disk therefore makes the contour integral continuous as the vertices move back. Each untranslated piece consequently has zero integral. Internal oriented edges cancel when the pieces are added, so every small triangle has zero integral. Morera's theorem proves the claim locally along the line. The continuity across $L$ is essential; holomorphicity on the two sides alone is insufficient.

Use the boundary form of Cauchy's integral theorem: the integral of a holomorphic function around the oriented boundary of a compact region on whose neighborhood it is holomorphic is zero, including annular regions with the inner boundary oppositely oriented. Apply it to $f(z)/(z-a)$ between the circles of radii $r$ and $\rho$, both around $a$. It givesas $\rho \downarrow 0$, by continuity. This proves the Cauchy integral formula.

For a bounded entire function with $|f|\le M$, apply that formula at variable points inside a fixed circle and differentiate its kernel under the integral. At its center $a$, this givesThe circle can have any radius because the function is entire. Letting $R\to \infty$ gives $f^{\prime }(a)=0$ at every $a$, so $f$ is constant. This proves Liouville's theorem.

Finally, if a nonconstant entire function omitted all points within distance $ϵ$ of some $w$, then $g=1/(f-w)$ would be entire and bounded by $1/ϵ$. Liouville's theorem would make $g$, and hence $f$, constant. Therefore every such neighborhood contains a value of $f$, proving the dense image of a nonconstant entire function property.

Use the following boundary version of Cauchy's integral theorem: if a function is holomorphic on a neighborhood of a compact region bounded by finitely many piecewise smooth simple closed curves, its integral around the positively oriented boundary is zero. On an annulus this means that its integrals around the two circles, both taken counterclockwise, are equal.

For $0<\rho <r$, the function $f(z)/(z-w)$ is holomorphic on a neighborhood of $\rho \le |z-w|\le r$. The theorem therefore givesAs $\rho \downarrow 0$, continuity of $f$ makes the integrand converge uniformly to $f(w)$. The right side tends to $f(w)$, proving Cauchy integral formula without assuming that formula in an intermediate step.

Let $I$ denote the requested boundary integral. Parametrize the unit circle by $z=e^{it}$. Since $dz=i{e}^{it}dt$,Taking complex conjugates and using $\overline{e^{it}}=e^{-it}$ givesThis is the conjugated Cauchy integral on a circle identity. The function $h(z)=f(z)/(1-\overline{a}z)$ is holomorphic on a neighborhood of the closed unit disc, because $|a|<1$. Applying Cauchy integral formula at zero gives $\overline{I}=h(0)=f(0)$. ThusThe complex conjugates in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.

Cauchy's integral theorem states that if a function is holomorphic on a simply connected open set, its integral around every closed piecewise-smooth path in that set is zero. A local form sufficient here is: if a function is holomorphic on an open set containing a triangle together with its interior, thenFor a general domain with holes, an arbitrary closed path need not have zero integral; the triangle or simple-connectedness hypothesis supplies the required interior.

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence

Cauchy's integral theorem says that if $f$ is holomorphic on an open set $U$ and a closed, piecewise smooth contour $\gamma$ is null-homotopic in $U$, thenIn particular, this holds for every closed contour when $U$ is simply connected, and for a simple closed contour whose boundary and enclosed region lie in $U$. For the later arguments it suffices to use the triangular version: the integral around a triangle is zero if the triangle and its interior lie in $U$ and $f$ is holomorphic there.

Both the holomorphicity condition and a suitable condition excluding holes inside the contour matter. Continuity alone, which is the starting hypothesis for this question, does not imply the theorem's conclusion; nor does holomorphicity on an arbitrary punctured open set imply zero integral around every loop.

Assume $q(z)=w^{\prime }(z)$ is single-valued and a holomorphic function between $C$ and a sufficiently large enclosing circle $|z|=R$. There are no fluid singularities in that region, so Cauchy's integral theorem permits contour deformation:Use the exterior Laurent series about zero, valid for $|z|$ larger than every finite singularity:It converges uniformly on the chosen circle, permitting termwise contour integration. For integer $n$, parameterization by $z=R{e}^{i\theta }$ givesOnly the $z^{-1}$ coefficient survives, provingThus if $b_1=\alpha +i\beta$, then $\Gamma =-2\pi \beta$ and $M=2\pi \alpha$. A uniform flow contributes a constant to $q$ and changes neither integral. The Laurent series here is an exterior expansion; it need not converge in a punctured neighborhood of zero if other singularities intervene.

Augustin-Louis Cauchy's importance lies in the interaction of foundations, methods and applications. He made the justification of limits, series and integrals a programme for teaching and research; developed complex analysis into a source of powerful general methods; and contributed to permutation groups, geometry and continuum mechanics. His work is best assessed through particular achievements and their limitations, rather than by treating every theorem now bearing his name as a modern statement he already possessed. His engineering training and subsequent mathematical career also make a rigid separation of pure and applied work historically unhelpful.^[ref]

The Cours d'analyse by Augustin-Louis Cauchy (1821) makes convergence a condition requiring argument. A series is studied through its partial sums; a limit describes a variable approaching a fixed value; an infinitesimal is a variable tending to zero, rather than simply a mysterious fixed quantity smaller than every positive number. The introduction explicitly challenges unrestricted transfers of rules from finite expressions to infinite ones. This is not an abandonment of calculation: it seeks to identify when calculations are valid.^[ref]

For example, the Cauchy criterion for series asks whether tails can all be made small:This is a modern formulation of the convergence requirement, with the completeness of the real numbers understood. Merely requiring $a_n\to 0$ is weaker. In the harmonic series, grouping the terms from $2^j+1$ to $2^{j+1}$ gives a block of at least $1/2$, so its tails cannot satisfy the criterion. In contrast, the root test compares a tail with a geometric series: if $\mathrm{limsup}|a_n{|}^{1/n}<q<1$, then eventually $|a_n|\le q^n$, which proves absolute convergence. Such arguments explain the usefulness of convergence tests instead of reducing Augustin-Louis Cauchy's contribution to a vocabulary of definitions.

His Résumé des leçons sur le calcul infinitésimal by Augustin-Louis Cauchy (1823) extends the programme to differentiation and integration. For a continuous function on a compact interval, an integral can be approached through sums of values times subinterval lengths. A modern justification uses uniform continuity: for a partition of mesh at most $\delta$, two choices of sample points give sums differing by at most $(b-a){\omega }_f(\delta )$, where the modulus of continuity ${\omega }_f(\delta )$ tends to zero. Refinement then establishes the common limit. This reconstruction makes clear why control of error, rather than a geometrical picture of area alone, is decisive. It does not attribute the later full Riemann integral theory for discontinuous functions to the 1823 text.^[ref]

The achievement has limits. The 1821 Sum theorem of Cauchy asserts continuity of sums of convergent series of continuous functions in terms that, under a pointwise reading, are insufficient. The original proof treats the tail as negligible without establishing the uniform control needed near the point concerned. A modern diagnostic is $f_n(x)=x^n$ on $[0,1]$: every $f_n$ is continuous, but its pointwise limit is zero for $x<1$ and one at $1$. Thus a proof must distinguish pointwise convergence from uniform convergence. Augustin-Louis Cauchy's use of variable and infinitesimal language deserves historical interpretation, but it does not license silently importing modern hypotheses into his printed theorem.^[ref]

In complex analysis, the Mémoire sur les intégrales définies, prises entre des limites imaginaires by Augustin-Louis Cauchy (1825) exemplifies a new productive setting for integration. A contour integral depends on a path in the complex plane; deforming that path and accounting for isolated singularities can evaluate otherwise difficult definite integrals. In modern form, Cauchy's integral theorem applies to a holomorphic function on a suitable domain and Cauchy integral formula states, for a circle enclosing $z$ and lying with its interior in that domain,The importance is structural: local values and derivatives are determined by data on a surrounding path. Historical proofs and hypotheses should be distinguished from later general forms of the theorem.^[ref]

A short modern consequence measures the power of this method. For an entire function bounded by $M$, the derivative version of Cauchy integral formula on a circle of radius $R$ gives $|f^{\prime }(z)|\le M/R$. Since $R$ is arbitrary, $f^{\prime }=0$, proving Liouville's theorem. If a nonconstant polynomial $P$ had no root, then $1/P$ would be an entire function, bounded outside a large disc by the leading term and inside the disc by compactness. Liouville's theorem would force it to be constant, a contradiction. This proves the fundamental theorem of algebra. The chain is an assessment of the reach of the methods, not a claim that Augustin-Louis Cauchy alone originated every named result.

Augustin-Louis Cauchy's algebra concerns concrete permutations as well as analytical formulas. His 1845 investigations developed the theory of substitutions; Cauchy theorem for groups says that a prime number dividing a finite group order occurs as an element order. One can appreciate its generality from a modern argument: rotate the coordinates of the tuples $(g_1,\dots ,g_p)$ satisfying $g_1\cdots g_p=e$. Rotation preserves this condition, since $g_2\cdots g_p{g}_1=g_1^{-1}(g_1\cdots g_p)g_1=e$. There are $|G{|}^{p-1}$ such tuples; nonfixed rotation orbits have $p$ elements, and fixed tuples correspond to $g^p=e$. When $p\mid |G|$, the number of fixed tuples is divisible by $p$, so besides $e$ there is an element of element order $p$. This is explanatory mathematics, not Augustin-Louis Cauchy's original proof. The dating itself needs care: Peter Michael Neumann's archival argument places the major burst of work in 1845 despite an 1844 date traditionally attached to a volume.^[ref]

Finally, Augustin-Louis Cauchy's work on elasticity shows the strength of making a local mathematical object out of physical interactions. Instead of describing only a force perpendicular to a surface, one studies the traction on a plane through a point, with its direction dependent on the plane's unit normal. In modern notation the relation is $\mathbf{t}(\mathbf{n})=\mathbf{\sigma }\mathbf{n}$, where the Cauchy stress tensor contains the necessary directional information. Together with balance of linear momentum, this yields equations for the internal motion of a body. The modern tensor notation is later explanatory language; the conceptual contribution is the reduction of the forces on arbitrary internal planes to a finite collection of local components.^[ref]

Augustin-Louis Cauchy's lasting contribution was a powerful research programme, not a finished foundation. His insistence on justified passage to a limit reshaped analysis, while complex integration, substitutions and local stresses opened methods whose value extended beyond their first applications. The inadequacy of a pointwise convergence hypothesis, the need to state domains in complex analysis, and the later construction of the real numbers show that nineteenth-century rigour remained a collective and revisable achievement. These qualifications strengthen an assessment of his influence by identifying what later mathematicians could extend and repair.

Fix real $y$ and use a rectangle with vertical sides at real coordinates $\pm R$. The function $e^{-z^2}$ is entire, so its closed contour integral is zero by Cauchy's integral theorem. The bottom horizontal side contributes ${\int }_{-R}^R{e}^{-x^2}dx$ and the top contributes the negative of ${\int }_{-R}^R{e}^{-(x+iy{)}^2}dx$. This identity applies for either sign of $y$, with the corresponding oriented rectangle.

On a vertical side, $z=\pm R+is$ with $|s|\le |y|$, andThe sum of the two vertical integrals in absolute value is at most $2|y|e^{-R^2+y^2}\to 0$. Both horizontal improper integrals converge absolutely, since $|e^{-(x+iy{)}^2}|=e^{y^2}{e}^{-x^2}$. Letting $R\to \infty$ therefore provesComplete the square in the Fourier transform:The shifted-integral identity applies after scaling by $\sqrt{2}$. The Gaussian integral is $\sqrt{2\pi }$: the square of $\int e^{-u^2}du$ becomes $2\pi {\int }_0^{\infty }{e}^{-r^2}rdr=\pi$ in polar coordinates. Thus the Gaussian Fourier transform yieldsThe eigenvalue reflects the unnormalized transform convention in this question.

Integrate $e^{iz}/z$ around the positively oriented indented upper-half-plane contour. The outer semicircle runs from $R$ to $-R$; the inner semicircle runs clockwise from $-r$ to $r$, excluding zero. The integrand is holomorphic inside this region, so Cauchy's integral theorem gives total integral zero.

The outer arc tends to zero by part (a) with $f(z)=1/z$ and $\lambda =1$. On the inner arc $z=r{e}^{i\theta }$, $\theta$ decreases from $\pi$ to zero, andThe two real segments combine asPassing to the limits gives $2iI-i\pi =0$, and hence the Dirichlet integralThis is an ordinary conditionally convergent improper integral: the integrand tends to one at zero, and integration by parts gives a tail bounded by $2/A$ beyond $A$. The clockwise indentation sign is essential.

Cauchy's integral theorem states that a holomorphic function on a simply connected domain has zero contour integral around every closed piecewise smooth curve in that domain. One proof begins with the Cauchy theorem for a triangle. Subdivide a triangle into four equal triangles. Their oriented boundary integrals sum to the original integral because internal edges cancel, so choose a nested subtriangle carrying at least one quarter of the preceding integral's modulus. At stage $n$, its perimeter and diameter are $P/2^n,d/2^n$, and the nested triangles converge to a point $z_0$. Complex differentiability gives $f(z)=f(z_0)+f^{\prime }(z_0)(z-z_0)+ϵ(z)(z-z_0)$, with $ϵ(z)\to 0$. The affine terms have primitives and integrate to zero. Therefore $|I_n|\le \sup |ϵ|Pd/4^n$, while $|I_0|\le 4^n|I_n|$. Letting $n\to \infty$ proves $I_0=0$. Triangulation cancels internal edges for polygonal contours; local primitives and deformation through the domain extend the result to the stated piecewise smooth curves. Simple connectedness ensures the necessary contractions do not pass through a missing point.

Apply the theorem to $e^{-a{z}^2}$ on the rectangle between heights zero and $\rho$, with vertical sides at $\mathrm{Re}z=\pm R$. On either vertical side,The modulus of each side integral is at most $|\rho |e^{-a{R}^2+a{\rho }^2}\to 0$. Both horizontal integrals converge absolutely. The two horizontal orientations are opposite, so the limit gives $I(\rho ,a)=I(0,a)$, also for negative $\rho$. This is Gaussian contour translation.

Now complete the square:The substitution $z=x-i\nu /(2a)$ moves the path to height $-\nu /(2a)$. Its Gaussian integral equals the real-axis integral by the preceding result. HenceThis also gives the Fourier transform of a Gaussian in the convention with no prefactor in the forward transform.

Write $f=u+iv$ and $z=x+iy$. For a positively oriented, piecewise smooth simple closed contour bounding a region where $u,v$ have continuous first partial derivatives, Green's theorem givesThe Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$, so both integrands vanish:This proves Cauchy's integral theorem under the standard regularity assumptions for Green's theorem. More general admissible contours follow by subdivision; every enclosed region must be free of singularities.

The function $e^{-z^2}$ is entire, so Cauchy's integral theorem makes its contour integral around the rectangle vanish. The lower horizontal integral tends to $\sqrt{\pi }$ by the Gaussian integral. On a vertical side $z=\pm R+it$, with $t$ between $0$ and $b$, its modulus is $e^{-R^2+t^2}\le e^{-R^2+b^2}$; hence both vertical integrals tend to zero. The upper side has the opposite orientation and contributesConsequently the integral in this expression is $\sqrt{\pi }{e}^{-b^2}$. Its imaginary part vanishes because $e^{-x^2}\sin (2bx)$ is odd, and its real part is twice the positive-half-line integral because the integrand there is even. ThereforeFor negative $b$ the rectangle lies below the real axis and the same cancellation holds; alternatively use the even dependence on $b$.

Let the common domain be an open set $U$. The uniform limit theorem makes the limit $f$ continuous. For every triangle whose closed interior lies in $U$, Cauchy's integral theorem gives ${\int }_{\partial T}{f}_n=0$. Part (ii) passes this identity to the limit, giving ${\int }_{\partial T}f=0$. Morera's theorem now proves that $f$ is holomorphic.

The same reasoning works for locally uniform convergence, since each triangle boundary is compact. No passage of derivatives through a limit has been assumed.

For $x>0$, the real and imaginary parts of the exponent divided by $x$ areThus constant-phase contours satisfy $u^3-3u{v}^2=c$. Through zero, $c=0$ gives the six rays $\theta =\pi /6+j\pi /3$. The three rays $\theta =\pi /6,5\pi /6,3\pi /2$ have decreasing real part of the exponent and are steepest descent contours; the alternating three are steepest ascent rays.

Through one, the phase is $c=1$. Its branch passing through that point can be writtenwith the cube root continuous from $t=1$. The exponent is $ix-xs$, so increasing $s$ from zero gives descent and decreasing $s$ gives ascent.

Deform the finite interval into the ray $t=e^{i\pi /6}r$, $0\le r<\infty$, followed by the reverse of the descent branch from one. The integrand is entire, so Cauchy's integral theorem permits the deformation.

For a precise connector at infinity, use $t=(a+iR{)}^{1/3}$, $0\le a\le 1$. Its length is $O(R^{-2/3})$ and the exponential has modulus $e^{-xR}$. Its contribution is therefore $O(R^{-2/3}{e}^{-xR})\to 0$. This avoids relying on an undifferentiated claim that every large circular arc decays.

Cauchy's integral theorem states that a holomorphic function on a simply connected open region has zero contour integral around every closed piecewise smooth contour in that region. More generally the same holds for a contour homologous to zero in its domain of holomorphy. Having no singularity on the curve alone is insufficient: $1/z$ has a nonzero contour integral around the unit circle.

Excise a small disc about zero and a thin slit along the negative real axis, using the orientations shown in the source: $C_1$ upwards, $C_3$ towards zero above the slit, $C_4$ clockwise around zero, and $C_5$ away from zero below the slit. On the resulting region $F$ is a single-valued holomorphic function; zero lies in the excluded disc and there are no poles in the region. Cauchy's integral theorem givesHere the paths on the cut mean limiting paths from its two sides. Apply Cauchy's integral theorem before taking those limits, when the entire contour lies inside the slit domain.

Cauchy's integral theorem states that a holomorphic function on a simply connected open region has zero contour integral around every closed piecewise smooth contour in that region. More generally the same holds for a contour homologous to zero in its domain of holomorphy. Having no singularity on the curve alone is insufficient: $1/z$ has a nonzero contour integral around the unit circle.