Integrate $$\int_C{\tan{z}\ dz}$$ $C$ is the parabola arc $y=x^2$ that connects the points $z=0$ and $z=1+i$.
This is what I've done so far:
- I know that $\tan{z}=\dfrac{\sin{z}}{\cos{z}}$
- And also, separating $\sin{z}$ and $\cos{z}$ in the form $z=U(x,y)+iV(x,y)$, that $$\sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}$$ $$\cos{z}=\cos{x}\cosh{y}-i\sin{x}\sinh{y}$$
So I used these formulas to separate $\tan(z)$ into that $U(x,y)+iV(x,y)$ form to check with the Cauchy-Riemman conditions if it is analytical (to integrate directly).
This is my result of doing this: $$\tan{z}=\dfrac{\sin{x}\cos{x}+i\sinh{y}\cosh{y}}{\cos^ 2{x}\cosh^2{y}+\sin^2{x}\sinh^2{y}}$$
As you can see, to check that using the Cauchy-Riemman conditions will be a tedious step to complete.
Is there an easier way to do this? Thanks in advance =)
P.S. If there is no easier way, what is the best way to derivate that fraction (as the CR conditions needs derivation to work)?
P.P.S. I'll be thankful if you tell me about possible mistakes I made 😉
Best Answer
$\tan(z)$ is analytic along the curve $C$. So we can choose another path $y=x$, which reduce the integral to
$$\int^1_0 \tan(1+i)x dx$$
You can then integrate using usual way (substitution).