Line integrals in the complex plane

Let $C$ be a piecewise-smooth curve in $\mathbb{C}$, parametrized as $z(t) = x(t) + iy(t)$ for $t\in[a,b]$, with $z(a)$ the starting point and $z(b)$ the ending point. For $f(z) = u(x,y) + iv(x,y)$ continuous on $C$, the complex line integral is

where: $z'(t) = x'(t) + iy'(t)$ is the tangent velocity of the parametrization, and the right-hand side is computed by ordinary real-variable calculus once $f(z(t))z'(t)$ is expanded into a single complex-valued function of the real variable $t$. Expanding into real and imaginary parts shows the same integral can be written as a pair of real line integrals,

but in practice the parametrization route $\int_a^b f(z(t))z'(t)\,dt$ is the standard computational tool, and it is the one used throughout this article. Three properties, inherited directly from the ordinary real integral, are used constantly: linearity, $\int_C[\alpha f + \beta g]\,dz = \alpha\int_C f\,dz + \beta\int_C g\,dz$; orientation reversal, $\int_{-C} f\,dz = -\int_C f\,dz$, where $-C$ is $C$ traversed backwards; and additivity over sub-arcs, $\int_C f\,dz = \int_{C_1} f\,dz + \int_{C_2} f\,dz$ when $C = C_1 + C_2$. A fourth tool, the ML-inequality, bounds the integral without computing it: if $|f(z)|\le M$ on $C$ and $C$ has length $L$, then

It follows from bounding $|f(z(t))z'(t)|$ pointwise and applying the triangle inequality for integrals, and it is the standard device for showing a contour integral vanishes in a limit — a technique reused heavily once residue calculus evaluates real integrals via contours later in the unit. A crucial early fact is that $\int_C f(z)\,dz$ can genuinely depend on the path $C$ joining two fixed endpoints, not just on the endpoints themselves — path-independence is a special property tied to analyticity, and it is developed fully in the very next topic, Cauchy's Integral Theorem.

Problem. Evaluate $\displaystyle\int_C z^2\, dz$ where $C$ is the straight-line segment from $z=0$ to $z=1+i$, and separately evaluate $\displaystyle\oint_C \frac{1}{z}\,dz$ where $C$ is the unit circle $|z|=1$ traversed once counter-clockwise.

Step 1. Parametrize the segment as $z(t) = t(1+i)$, $t\in[0,1]$, so $z(0)=0$, $z(1)=1+i$, and $z'(t)=1+i$. Then $z(t)^2 = t^2(1+i)^2 = t^2(2i) = 2it^2$, since $(1+i)^2 = 1+2i-1 = 2i$.

Step 2. Apply the reduction formula from the derivation: $\int_C z^2\,dz = \int_0^1 2it^2\cdot(1+i)\,dt = 2i(1+i)\int_0^1 t^2\,dt = 2i(1+i)\cdot\tfrac13 = \tfrac{2i(1+i)}{3} = \tfrac{2i-2}{3} = -\tfrac23 + \tfrac23 i$.

Step 3. Sanity-check using the antiderivative. Because $z^2$ is analytic everywhere, this integral can also be computed as $\left[\tfrac{z^3}{3}\right]_0^{1+i} = \tfrac{(1+i)^3}{3}$, and $(1+i)^3=(1+i)(2i)=2i-2$, giving the same $-\tfrac23+\tfrac23 i$ — consistent with $z^2$ possessing an antiderivative on all of $\mathbb{C}$.

Step 4. Now the circle: parametrize $C$ by $z(t)=e^{it}$, $t\in[0,2\pi]$, so $z'(t)=ie^{it}$, and $\tfrac{1}{z(t)}\cdot z'(t) = \tfrac{ie^{it}}{e^{it}} = i$, a constant. So $\oint_C \tfrac1z\,dz = \int_0^{2\pi} i\,dt = 2\pi i$.

The answers are: $\int_C z^2\,dz = -\dfrac23+\dfrac23 i$ on the segment to $1+i$, and $\displaystyle\oint_{|z|=1}\dfrac{dz}{z}=2\pi i$.

Competitive stretch

Bound $\left|\int_C z^2\,dz\right|$ for the same segment using only the ML-inequality, without computing the integral. On $C$, $|z|\le\sqrt2$ (the farthest point is $1+i$), so $|f(z)|=|z|^2\le M=2$; the segment has length $L=|1+i|=\sqrt2$. The ML-inequality gives $\left|\int_C z^2\,dz\right|\le ML = 2\sqrt2 \approx 2.83$. Checking against the exact value: $\left|-\tfrac23+\tfrac23 i\right| = \tfrac23\sqrt2 \approx 0.94$, comfortably below the bound — as it must be, since ML only ever gives an upper estimate, never the exact answer.

• Treating $dz$ like $dx$. Always substitute $z(t)$ into $f$ and multiply by $z'(t)\,dt$, never just $dt$ — dropping $z'(t)$ is the single most common arithmetic slip.
• Assuming every contour integral is path-independent. The unit-circle example shows a non-analytic-at-a-point integrand giving a nonzero closed-loop integral; path-independence is guaranteed only under Cauchy's Integral Theorem's hypotheses, covered next.
• Sign errors from orientation. Reversing the direction of traversal flips the sign of the whole integral; fix and state the direction (usually counter-clockwise for closed contours) before parametrizing.
• Misusing ML as an equality. The ML-inequality gives only an upper bound $|\int_C f\,dz|\le ML$, useful for showing an integral is small or vanishes in a limit — never for computing the exact value, as the competitive stretch above makes explicit.

Ask the tutor: if parametrizing curves other than straight segments and full circles (semicircular arcs, quarter-arcs, or paths made of several joined pieces) feels unfamiliar, ask the AI tutor to walk through a few more contour setups before moving on to Cauchy's Integral Theorem.