Simple Line Integral in Complex Numbers

[Tsunoyukami]

"Simple" Line Integral in Complex Numbers

If anyone could please double-check my final result for this question it would be greatly appreciated. Rather than write out each step explicitly, I'll explain my approach and write out only the most important parts.

"[E]valuate the given integral...

10. $\int_{\gamma} (z + \frac{1}{z}) dz$,

where $\gamma$ is any curve in $Im z > 0$ joining $-4+i$ to $6+2i$." (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 117)


In general we can find a function $F(z)$ such that $F'(z) = f(z)$. Then $\int_{\gamma} f(z) dz = F(b) - F(a)$ where a and b are the staring and end points, respectively.

This question has $f(z) = (z + \frac{1}{z})$; therefore, we can find $F(z) = \frac{z^{2}}{2} + log(z)$. Setting $a = -4 + i$ and $b = 6 +2i$ we find the following:

$$\int_{\gamma} (z + \frac{1}{z}) dz = \frac{(6+2i)^{2}}{2} + log(6+2i) - \left[\frac{(-4+i)^{2}}{2} + log(-4+1)\right]$$

I will skip my algebra, but this is my final result.

$$\int_{\gamma} (z + \frac{1}{z}) dz = 16 - \frac{15}{2} + ln(40) - ln(17) + i \left[16 + tan^{-1}(\frac{1}{3}) - tan^{-1}(\frac{1}{4})\right]$$


I would appreciate it greatly if someone could check and verify my answer for me. Thanks a bunch!

Note: Here I used the fact that $log(z) = ln|z| + iarg(z)$.

 

[Millennial]

To check your answer, use the definition of a contour integral and use the parametrization \displaystyle z = t + \frac{i}{5}(t+4) from t = -4 to t=6 to integrate the function given to you. If your answer is correct, it must match the answer obtained from this integral.

 

[Tsunoyukami]

Just to be sure, my result should match the integral obtained for any curve connecting the start and end point in my original post that lies entirely in I am z > 0, correct? If this is so I could use any parametrization for $\gamma$ that I thought would be "easiest", correct?

I am unfamiliar with the parametrization you have suggested; does this parametrize a line segment? How did you arrive at this parametrization? Just curious to see other useful methods of parametrizing curves. :)


EDIT: Unless my above idea is incorrect (ie. that any curve connecting these two points in this domain should yield the same result) I've been getting different answers. I used the parametrization provided by Millennial as well as a parametrization connecting the two points in straight line (ie. $\gamma = tz_{2} + (1-t)z_{1}$) and computed different results using wolfram alpha, none of which agreed with my original result. I have no issue accepting that my original result is incorrect, but am confused by the fact that the two line integrals yielded different results...

Let me double-check what you meant by contour integral, as I am unfamiliar with the terminology.

$$\int_{\gamma} f(z) dz = \int_{a}^{b} f(\gamma(t)) \cdot \gamma '(t) dt$$

Is this what you meant by contour integral?


For the sake of completeness and because I have yet to determine whether or not my solution is correct I will post my algebra; hopefully someone can point out where I've gone wrong if I have done so:

$$\int_{\gamma} f(z) dz = \int_{\gamma} \left( z + \frac{1}{z} \right) dz$$

Then I find a function $F(z)$ that satisfies the condition $F'(z) = f(z) = z + \frac{1}{z}$; this function is $F(z) = \frac{z^{2}}{2} + log(z)$. I can then write:

$$\int_{\gamma} \left( z + \frac{1}{z} \right) dz = \int_{\gamma} f(z) dz = \int_{\gamma} F'(z) dz = F(b) - F(a)$$

Using the $F(z)$ found above and the points $a = (-4 +i)$ and $b = (6 +2i)$:

$$F(b) - F(a) = \left( \frac{b^{2}}{2} + log(b) \right) - \left( \frac{a^{2}}{2} + log(a) \right) = \frac{(6 +2i)^{2}}{2} + log(6+2i) - \frac{(-4+i)^{2}}{2} - log(-4+i)$$

$$F(b) - F(a) = \frac{36 +24i -4}{2} + ln\left|6+2i\right| + iarg(6+2i) - \frac{16-8i-1}{2} - ln\left|i -4\right| - iarg(i-4)$$

$$F(b) - F(a) = \frac{32-24i}{2} + ln\left( (36+4)^{\frac{1}{2}}\right) + iarctan(2/6)...$$

Now at this point I realize I forgot to take the square root while computing the modulus required by the complex representation of the logarithm function, so I'm going to try plugging these numbers into wolfram to see if I get the same result...I guess sometimes just working through it repeatedly is the trick! (I'll be back if it doesn't work out...)