# Curves in the Complex Plane

Suppose the continuous real-valued functions $x = x(t),$ $y = y(t),$ $a \leq t \leq b,$ are parametric equations of a curve $C$ in the complex plane. If we use these equations as the real and imaginary parts in $z = x+iy$, we can describe the points $z$ on $C$ by means of a complex-valued function of a real variable $t$ called a parametrization of $C$: \begin{equation}\label{parcurve} z(t) = x(t) + i y(t), \quad a\leq t\leq b. \end{equation}

The point $z(a) = x(a) + i y(a)$ or $z_0 = (x(a), y(a))$ is called the initial point of $C$ and $z(b) = x(b) + iy(b)$ or $z_1 = (x(b), y(b))$ is its terminal point. The expression $z(t) = x(t) + iy(t)$ could also be interpreted as a two-dimensional vector function. Consequently, $z(a)$ and $z(b)$ can be interpreted as position vectors. As $t$ varies from $t = a$ to $t = b$ we can envision the curve $C$ being traced out by the moving arrowhead of $z(t)$. This can be appreciated in the following applet with $0 \leq t\leq 1$.

Press Start to animate. You can move the points to change the curve.

For example, the parametric equations $x = \cos t,$ $y = \sin t,$ $0 \leq t \leq 2\pi,$ describe a unit circle centered at the origin. A parametrization of this circle is $z(t) = \cos t + i \sin t,$ or $z(t) = e^{it},$ $0 \leq t \leq 2\pi$.

Press Start to animate.

## Contours

The notions of curves in the complex plane that are smooth, piecewise smooth, simple, closed, and simple closed are easily formulated in terms of the vector function (\ref{parcurve}). Suppose the derivative of (\ref{parcurve}) is $z'(t) = x'(t) + iy'(t).$ We say a curve $C$ in the complex plane is smooth if $z'(t)$ is continuous and never zero in the interval $a \leq t \leq b.$ As shown in Figure 2, since the vector $z'(t)$ is not zero at any point $P$ on $C$, the vector $z'(t)$ is tangent to $C$ at $P$. Thus, a smooth curve has a continuously turning tangent; or in other words, a smooth curve can have no sharp corners or cusps. See Figure 2. Figure 1: $z'(t)=x'(t) + iy(t)$ as a tangent vector. Figure 2: Curve $C$ is not smooth since it has a cusp.

A piecewise smooth curve $C$ has a continuously turning tangent, except possibly at the points where the component smooth curves $C_1, C_2, \ldots, C_n$ are joined together.

A curve $C$ in the complex plane is said to be a simple if $z(t_1) \neq z(t_2)$ for $t_1 \neq t_2,$ except possibly for $t = a$ and $t = b.$ $C$ is a closed curve if $z(a) = z(b).$

$C$ is a simple closed curve if $z(t_1)\neq z(t_2)$ for $t_1\neq t_2$ and $z(a) = z(b).$

In complex analysis, a piecewise smooth curve $C$ is called a contour or path. We define the positive direction on a contour $C$ to be the direction on the curve corresponding to increasing values of the parameter $t$. It is also said that the curve $C$ has positive orientation. In the case of a simple closed contour $C$, the positive direction corresponds to the counterclockwise direction. For example, the circle $z(t) = e^{it}$, $0 \leq t \leq 2\pi$, has positive orientation. The negative direction on a contour $C$ is the direction opposite the positive direction. If $C$ has an orientation, the opposite curve, that is, a curve with opposite orientation, is denoted by $−C$. On a simple closed curve, the negative direction corresponds to the clockwise direction. For instance, the circle $z(t) = e^{-it}$, $0 \leq t \leq 2\pi$, has negative orientation.
Press Start to animate. You can change the direction of $C$.
Exercise: There is no unique parametrization for a contour $C$. You should verify that \begin{eqnarray*} z(t)&=&e^{it} =\cos t+i\sin t,\; 0\leq t\leq 2\pi \\ z(t)&=& e^{2\pi it} =\cos \left(2 \pi t\right) +i \sin \left(2 \pi t\right),\; 0\leq t\leq 1\\ z(t)&=&e^{\pi/2 it} =\cos \left( \frac{\pi}{2} t\right)+i \sin \left( \frac{\pi}{2} t\right),\; 0\leq t \leq 4 \end{eqnarray*} are all parametrizations, oriented in the positive direction, for the unit circle $|z| = 1$.