Cauchy integral problem: can this answer be simplified further?

[redshift]

The question calls for using Cauchy's integral formula to compute the integral for Int.c z/[(z-1)(z-3i)] dz, assuming C is the loop |z-1|=3.
Taking z = 1 and f(z) = z/(z-3i), I came up with (2pi*i)/(1-3i), which seems like it could be simplified, but I'm not sure how.

 

[xepma]

Multiply the numerator and denominator with the complex conjugate of 1-3i = 1+3i. Than you will get rid of the 'i' in the denominator.

 

[wais]

It is possible to simplify the answer further using the properties of complex numbers. First, we can rewrite the denominator as (z-1)(z-3i) = z^2 - (3+1i)z + 3i. Then, we can use the quadratic formula to find the roots of this polynomial, which are z = 3i and z = 1 - 2i.

Next, we can rewrite the integral as Int.c z/[(z-1)(z-3i)] dz = Int.c z/(z-3i) dz - Int.c z/(z-1) dz. Using Cauchy's integral formula, we can evaluate each of these integrals separately.

For the first integral, we can use the point z = 3i and the function f(z) = z/(z-3i) to get (2pi*i)/(1-3i).

For the second integral, we can use the point z = 1 - 2i and the function f(z) = z/(z-1) to get (2pi*i)/(-2i).

Finally, we can combine these two results to get the simplified answer of (pi*i)/(1+i).

In summary, by using the roots of the denominator and evaluating the integral separately, we can simplify the answer to (pi*i)/(1+i).