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I'm sorry if I sound too ignorant, I don't have a high level of knowledge in math.

Interested in learning to integrate through other paths besides the real line in the complex plane, I searched how to do so in the internet. The most understandable "proof" of the "Line Integration Formula" was the following:

$C$ is a contour defined by $P(t)=R(t)+iI(t)$, $a<t<b$

$∫f(z)dz$ (through $C$) $=∫f(z(t))z'(t)dt$ (from $a$ to $b$)

since $z$ is in function of $t$, and $z'(t)=dz/dt$, so $dz$=$z'(t)dt$

The latter confuses me in the first step, when we state that the Integration of $C$ in $f$ $=∫f(z)dz$ (through $C$).

With the $∫f(z)$, I can imagine the sum of many (infinite) inputs of $C$ in $f$. But what is the meaning of $dz$?

In "real" integrals, $dx$ can be seen as $(b-a)/m$, where $m$ is the number of inputs we are adding, but I struggle when trying to see $dz$ in the same way since I can´t find the parallel of that $(b-a)$ in complex integrals. My first intuition was to think of $(b-a)$ as the length of the curve, but the "Line Integration Formula" seems to pay no attention on the length of the contour it is integrating.

Is there a way to approximate a complex integral in a similar way to the "real" integrals? I believe such approximation, if it exists, might give me an insight on the meaning of $dz$.

I think that my question is hard to explain, so if there is something that is confusing I would be happy to clarify.

I would really appreciate any explanation or insight!

Sam
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2 Answers2

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Suppose we parametrize the contour $C$ as $\gamma(t),\; a \le t \le b$. The contour integral $\int_C f(z)\; dz$ is the limit as $n \to \infty$ of "Riemann sums" $$ \sum_{j=1}^n f(z_j)\; \Delta z_j$$ where $z_j = \gamma(t_j)$ and $\Delta z_j = z_j - z_{j-1}$, $a = t_0 < t_1 < \ldots < t_n = b$, if $\max(t_{j} - t_{j-1}) \to 0$.

Robert Israel
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Before explaining about dz, take L as a Load/Force to be a function of x, say L= 4x. If we now care to give this a meaning we can say that this function represents a man moving in a straight line a distance x and as he does, he has to push against a force y which changes as he moves and described as 4x. Now, if we had to look at the work done by the man, we have to multiply the Force he suffers by some distance so in this case we first take a small distance dx. So the work done by the man is given by Work = L dx = 4x dx. From this we can find the total work done.

The above problem was a man walking in a line so really it is a one dimensional problem. If we have to have the man walking in an area of two dimensions then we have to use the expression z=x+jy. and so dz= dx+jdy

dz in this case may be looked as a small distance in a two dimensional plane when the man moves in the dx and the dy direction. Now as before if a man is made to face a force of F(z) as the walks on a two dimensional distance z= x + jy, then he too would be subjected to having to do work and so one can look upon F(z)dz exactly the same as one looked at P(x) dx.

Since z=x+jy is a position vector then any function F(z) will be a vector so so here we have a situation where the poor man is subjected to a Load or a Force which is not only changing in magnitude but also in direction as he walks the plane x+jy. All one needs to do is to imagine the Load. Force F(z) acting at location z and as z changes the change is dz. If the man is drunk and walks through a contour then the change of direction dz is tangential to the curve he walks so F(z)dz is exactly the same as P(x)dx. which, if taken between two points in the trajectory he walks, could be interpreted as the work done by a drunken man walking a curve on the flat plane a distance z=x+jy from the origin and facing a force F(z) which is always changing in magnitude and direction depending on the distance he is located from the origin of his walk. This integrated between limits of the trajectory the man walks is the work done by the man.

Carmel Pule'
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