- #1
Tsunoyukami
- 215
- 11
"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)##.
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)##.