4.3: Fundamental Theorem for Complex Line Integrals

Last updated
Save as PDF

Page ID: 6482

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

This is exactly analogous to the fundamental theorem of calculus.

Theorem \(\PageIndex{1}\): Fundamental Theorem of Complex Line Integrals

If \(f(z)\) is a complex analytic function on an open region \(A\) and \(\gamma\) is a curve in \(A\) from \(z_0\) to \(z_1\) then

\[\int_{\gamma} f'(z) \ dz = f(z_1) - f(z_0). \nonumber \]

Proof

This is an application of the chain rule. We have

\[\dfrac{df(\gamma (t))}{dt} = f'(\gamma (t)) \gamma '(t). \nonumber \]

\[\int_{\gamma} f'(z) \ dz = \int_{a}^{b} f'(\gamma (t)) \gamma '(t)\ dt = \int_{a}^{b} \dfrac{df(\gamma (t))}{dt} \ dt = f(\gamma (t)) \vert_{a}^{b} = f(z_1) - f(z_0) \nonumber \]

Another equivalent way to state the fundamental theorem is: if \(f\) has an antiderivative \(F\), i.e. \(F' = f\) then

\[\int_{\gamma} f(z)\ dz = F(z_1) - F(z_0). \nonumber \]

Example \(\PageIndex{1}\)

Redo \(\int_{\gamma} z^2\ dz\), with \(\gamma\) the straight line from 0 to \(1 + i\).

Solution

We can check by inspection that \(z^2\) has an antiderivative \(F(z) = z^3/3\). Therefore the fundamental theorem implies

\[\int_{\gamma} z^2\ dz = \left. \dfrac{z^3}{3} \right\vert_{0}^{1 + i} = \dfrac{(1 + i)^3}{3} = \dfrac{2i(1 + i)}{3}. \nonumber \]

Example \(\PageIndex{2}\)

Redo \(\int_{\gamma} z^2 \ dz\), with \(\gamma\) the unit circle.

Solution

Again, since \(z^2\) had antiderivative \(z^3/3\) we can evaluate the integral by plugging the end-points of \(\gamma\) into the \(z^3/3\). Since the endpoints are the same the resulting difference will be 0!