If f(x)dx is a rectangle with height f(x) and width dx, what is f(z)dz in complex analysis I am trying to intuitively understand the multiplication $f(z)dz$ in complex analysis. For instance, $f(x)dx$, we are all aware, is a rectangle with height $f(x)$ and width $dx$ so its multiplication is an area of this rectangle. Is there a similar way to visualize $f(z)dz$ also?
 A: You can interpret the function $f$ as a "complex force field" defined in the region $\Omega$; the function value $f(z)$ is then the force vector attached to the point $z$. On the other hand $dz$ represents an infinitesimal movement from $z$ to $z+dz$. The product $f(z)\>dz$ is then the "complex work" done when a mass point moves under the influence of $f$ from $z$ to $z+dz$.
Note that the notion of line integral picks up this intuition:
$$\int_\gamma f(z)\>dz=\lim_\ldots \sum_{k=1}^N f(z_k)(z_k-z_{k-1})\ ,$$
where
$$\gamma:\quad t\mapsto z(t)\qquad(a\leq t\leq b)$$
is a curve in $\Omega$, and $z_k=z(t_k)$ for a partition $a=t_0<t_1<\ldots<t_N=b$.
A: When you are taking an integral with respect to $z$ you are integrating along curves in the complex plane. Let us assume there is just one continuous curve, without loss of generality. Then the curve can be parameterized in terms of a single real variable, say $t$, so there are real valued functions $p(t)$ and $q(t)$ such that $z=p(t)+iq(t)$. We may assume that the parametrization has domain $[0,1]$. The integral becomes
$$\int f(p(t)+iq(t))d(p(t)+iq(t)) =\int_0^1 f(p(t)+iq(t))(p'(t)+iq'(t))dt$$
This allows you to interpret dt analogously to the width of a rectangle, but the base of the rectangle is twisted along the curve of integration. The integrand may be interpreted as the height of the rectangle, even though it is complex: it splits up into a "real height" and an "imaginary height".
