# The complex integral

This applet is a little less intuitive than previous ones; it tries to display the process of complex integration. If you draw a curve on the left-hand side of the grid, the applet will integrate the function f(z) along your curve on the right side of the grid. For instance, if f(z) = z and you drag the mouse from 1 to 2i, the pointer on the right-hand grid will be at the value of the integral of z from 1 to 2i. The value of the integral always starts at 0.

If the mouse is at a location z, and you drag the mouse by small amount dz, then the integral gets changed by an amount f(z) dz. The cyan and green lines indicate the direction the integral would move by if you moved z rightward or upward respectively; they represent the complex numbers f(z) and i f(z) respectively.

When you integrate f on a closed loop, you may or may not get 0 as the total integral. It depends a lot on what f is, and what the loop is. If f has an anti-derivative on the loop, then the integral is guaranteed to be 0 by the fundamental theorem of calculus. If f is analytic on the loop and inside the loop, then Cauchy's theorem guarantees that the integral is 0. In all other cases it is very unlikely that the integral vanishes completely.

Notes on selected functions:
• f(z) = 1. If you integrate the function 1 from a to b, you should just get b-a. Check that the above applet agrees with this reasoning. What is the integral of a closed loop?
• f(z) = 1+i. What happens when you drag the mouse rightward? upward? downward? leftward? What is the integral of a closed loop?
• f(z) = z. Now the function is changing with z. Note that the effect of extending the curve by dz now depends on the current location z. Nevertheless, the integral of a closed loop remains zero. Can you explain why?
• f(z) = zbar. Integrals of this function on closed loops are almost never zero. Can this function have an anti-derivative?
• f(z) = 1/z. Integrate this function on a closed loop which doesn't circle the origin. Now integrate it on a loop circling the origin anti-clockwise. Now integrate it on a loop circling the origin clockwise. What happens? Can you explain this by the theorems in the textbook? Can 1/z have an anti-derivative on the punctured plane C\{0}? What about if we only want an anti-derivative on the complex plane with 0 and the negative real axis removed?

• Previous applet: The complex derivative
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