Complex integrals are taken along curves in the plane, just like the line integrals of vector calculus. Before integrating anything we need the curves themselves, written as $z(t) = x(t) + iy(t)$. Two shapes do almost all the work in this course: segments and circles, and exponential form makes circles especially short to write.
Key idea
A contour is a piecewise smooth curve $z(t)$, $a \le t \le b$, with a direction (from $z(a)$ to $z(b)$).
• Segment from $z_1$ to $z_2$: $z(t) = z_1 + t(z_2 - z_1)$, $0 \le t \le 1$, with $z'(t) = z_2 - z_1$.
• Circle $|z - a| = r$, once counterclockwise: $z(t) = a + re^{it}$, $0 \le t \le 2\pi$, with $z'(t) = ire^{it}$.
• The same circle clockwise: $z(t) = a + re^{-it}$.
The length is $\int_a^b |z'(t)|\,dt$; for the circle it is $2\pi r$. The symbol $\oint_C$ means a closed contour traced once counterclockwise unless stated otherwise.
• Segment from $z_1$ to $z_2$: $z(t) = z_1 + t(z_2 - z_1)$, $0 \le t \le 1$, with $z'(t) = z_2 - z_1$.
• Circle $|z - a| = r$, once counterclockwise: $z(t) = a + re^{it}$, $0 \le t \le 2\pi$, with $z'(t) = ire^{it}$.
• The same circle clockwise: $z(t) = a + re^{-it}$.
The length is $\int_a^b |z'(t)|\,dt$; for the circle it is $2\pi r$. The symbol $\oint_C$ means a closed contour traced once counterclockwise unless stated otherwise.
Why it works
$e^{it} = \cos t + i\sin t$ is the point at angle $t$ on the unit circle, so $re^{it}$ is on the circle of radius $r$ and $a + re^{it}$ on the circle centered at $a$. As $t$ grows, the angle grows: counterclockwise. Replacing $t$ by $-t$ reverses the direction. The derivative $z'(t) = ire^{it}$ is the radius vector $re^{it}$ turned a quarter turn by the factor $i$: tangent to the circle, as a velocity should be, with speed $|z'(t)| = r$.
Worked examples
Worked example: A segment
Parametrize the segment from $1 + i$ to $3 - 2i$.
- The difference is $(3 - 2i) - (1 + i) = 2 - 3i$.
- $z(t) = (1 + i) + t(2 - 3i)$, $0 \le t \le 1$; $z'(t) = 2 - 3i$.
- Check: $z(0) = 1 + i$ and $z(1) = 3 - 2i$. Its length is $|2 - 3i| = \sqrt{13}$.
Answer: $z(t) = 1 + i + t(2 - 3i)$
Worked example: A point on a circle
For $z(t) = 2i + 3e^{it}$, find $z\left(\frac{\pi}{2}\right)$ and $z'\left(\frac{\pi}{2}\right)$.
- $e^{i\pi/2} = i$, so $z\left(\frac{\pi}{2}\right) = 2i + 3i = 5i$: the top of the circle centered at $2i$ with radius $3$.
- $z'(t) = 3ie^{it}$, so $z'\left(\frac{\pi}{2}\right) = 3i \cdot i = -3$: at the top, the motion is to the left, which is counterclockwise.
Answer: $5i$ and $-3$
Worked example: A closed path in two pieces
Describe the boundary of the upper half-disk of radius $2$, traced counterclockwise, and find its length.
- Piece 1: the segment from $-2$ to $2$, $z(t) = t$ for $-2 \le t \le 2$.
- Piece 2: the upper half circle from $2$ back to $-2$, $z(t) = 2e^{it}$ for $0 \le t \le \pi$.
- Length: $4 + 2\pi$. This shape is used later to evaluate integrals over the whole real line.
Answer: segment plus half circle; length $4 + 2\pi$
Common mistake
Using the counterclockwise formula for a clockwise path. $a + re^{it}$ always runs counterclockwise. If a problem asks for clockwise (or "negatively oriented"), use $a + re^{-it}$, or keep $e^{it}$ and run $t$ from $2\pi$ down to $0$. Reversing the direction flips the sign of every contour integral, so this is not a cosmetic choice. Check by computing $z'(0)$: for counterclockwise motion it points straight up from the rightmost point.