# 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 1, 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 for a counterexample.

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 $|z| = 1.$