Find a parametric equation to solve a line integral

Question

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?

Answer 1

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.

Answer 2

Hint :

$$z(t)=(b-a)t+a$$ $$ t\in [0,1]$$

So $$z(0)=a$$

$$z(1)=b$$

Answer 3

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

$$ln|5 + 6i| - ln|1 - 5i|$$

Answer 4

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.