Find a parametric equation to solve a line integral Find 
$ \int_C \frac{dz}{z} $ from $1 - 5i$ to $5 + 6i$
I want to use 
$$\int_C f(Z) dz = \int_a^b f[z(t)]z'(t)dt$$
My guess is that I need to find $z(t)$ such that 
$$z(a) = 1 - 5i$$
$$z(b) = 5 + 6i$$
But how?
 A: I found a displacement vector $<4, 11>$ such that $<1, -5> + <4, 11> = <5, 6>$
Now I can define a parametric line in terms of these vectors $<1, -5> + t<4, 11> = <1 + 4t, -5 + 11t>$ which is equivalent to :
$$z(t) = 1 + 5t + i(-5 + 11t)$$
$$0 \leq t \leq 1$$ 
since
$$z(0) = 1 - 5i$$
$$z(1) = 5 + 6i$$
Then we deal with the integral:
$$\int_c \frac{1}{z} = \int_0^1 \frac{4 + 11i}{1 - 5i + t\left(4 + 11i\right)}dt  \hspace{1cm} 0 \leq t \leq 1$$
which gives the result
$$ln|5 + 6i| - ln|1 - 5i|$$
A: You can actually use any path from $a$ to $b$.  You should get the same answer as long as the path does not include the origin.  This is the Principle of Path Independence.  It follows from the Cauchy-Goursat Theorem.
A: Hint : 
$$z(t)=(b-a)t+a$$ $$  t\in [0,1]$$
So $$z(0)=a$$
$$z(1)=b$$
A: Notice that your answer is equal $ln|b| - ln|a|$.  Since the integrand $\frac{1}{z}$ is continuous everywhere except $0$ and the path doesn't go near $0$ you can get the same answer using the Fundamental Theorem for Line Integrals.
