Complex line integrals in increasing directions The problem I am stuck on is :Evaluate $\displaystyle\int\frac{dz}{z^2+4}$ along the line $x+y=1$ in the direction of increasing x ..... Nothing I have learned in my independent study of this subject seems to help me here.......
 A: You have two options here. Without giving away too much, I'll describe both of them.
One is to parametrerize the contour with something like $x = t$, $y = 1-t$ for $t \in (-\infty,\infty)$. Then, $z = x+iy = t+i(1-t)$, $dz = (1-i)dt$, and you can evaluate the contour integral as a simple one variable integral.
The other option is to let $\Gamma_R$ be a contour consisting of a straight line $L_R$ from $z = -R+i(1+R)$ to $z = R+i(1-R)$, and then a counterclockwise circular arc $C_R$ from $z = R+i(1-R)$ back to $z = -R+i(1+R)$. (To specify a precise curve, we should say that the center of this circular arc is $z = i$, but that doesn't matter too much.) By the residue theorem, $\displaystyle\int_{L_R}\dfrac{dz}{z^2+4} + \displaystyle\int_{C_R}\dfrac{dz}{z^2+4} = \displaystyle\oint_{\Gamma_R}\dfrac{dz}{z^2+4} = 2\pi i \text{Res}\left(\dfrac{1}{z^2+4};2i\right)$ as long as $R$ is large enough to enclose the pole at $z = 2i$. You can show that as $R \to \infty$, the integral $\displaystyle\int_{C_R}\dfrac{dz}{z^2+4}$ tends to $0$, while the integral $\displaystyle\int_{L_R}\dfrac{dz}{z^2+4}$ tends to the line integral $\displaystyle\int\dfrac{dz}{z^2+4}$ along the infinite line $x+y = 1$. Thus, the answer is simply $2\pi i \text{Res}\left(\dfrac{1}{z^2+4};2i\right)$, which you can easily compute.
Let me know if you need more information about either method.
A: Let $I$ be the path integral given by 
$$I=\int_C \frac{1}{z^2+4}dz$$
where $C$ is the straight line $y=1-x$ and goes from $(-\infty,\infty)$ to $(\infty,-\infty)$.
Now, we truncate $C$ to $C_T$ so that $C_T$ forms the diameter of a semi-circle $C_R$ of radius $R$ and center $(1/2,1/2)$ and lies in the half-plane $y\ge 1-x$.  
Next, we evaluate the closed-contour integral 
$$\oint_{C_T+C_R}\frac{1}{z^2+4}dz$$
From the residue theorem we have
$$\oint_{C_T+C_R}\frac{1}{z^2+4}dz=2\pi i \frac{1}{4i}=\frac{\pi}{2}$$
since only the pole at $z=2i$ is enclosed.  Observe that as $R\to \infty$ the integral over $C_R$ behaves asymptotically as $O(R^{-1})$ and therefore goes to $0$, while the integral over $C_T$ goes to $I$.  
Thus, we find 
$$\bbox[5px,border:2px solid #C0A000]{\int_C \frac{1}{z^2+4}dz=\frac{\pi}{2}}$$
