I was given the following question in a small exam, which i had to answer within 5 minutes: Let $a,b \in \mathbb{C}$. Let $\gamma$ be the parameterization of the line between $a%$ and $b$. Compute: $$ \int_\gamma \left\lvert z-\mathcal{i} \right\rvert ^2\mathbb{d}z$$
I tried to calculate the integral with cartesian coordinates and the parameterization $\gamma :t\mapsto a+t(b-a)$, using the definition of a line integral: $$=\int_0^1|a+t(b-a)-\mathcal{i}|^2(b-a)\mathbb{d}t \ =\int_0^1\Bigl(\bigl(\Re(a)+t \left(\Re(b)-\Re(a)\right)\bigl)^2+ \bigl(\Im(a)+t\left(\Im(b)-\Im(a)\right) - 1\bigl)^2\Bigl)(b-a)$$
Which evaluates to
$$\frac{1}{3}(b-a)\bigl(\lvert a\rvert^2+ \lvert b\rvert^2+ \Re(a)\Re(b)+\Im(a)\Im(b)\bigl) + (a-b)\bigl(\Im(a)+\Im(b)-1\bigl)$$
While it is not hard to calculate this by multiplying and integrating the resulting polynomial, it is rather laborious and it took me far longer that 5 minutes. This leads me to believe that there is a more efficient way to do this. The only similar integrals that I was able to find on here were along a circle, where polar coordinates obviously facilitate the computation. Since the integrand is not holomorphic, I am not able to use Cauchy's integral formula either.
Is there a faster way to compute this, or am I just too slow using the method that I described?
Its easier to compute a general case then check parameters: $$\int_0^1 |A+tB|^2 Cdt = C\int_0^1|A|^2+2\Re \big(tA\overline B\big) + t^2|B|^2dt = C(|A|^2+\Re (A\overline B) + \frac13|B|^2)$$ now your integral is $$ A=a-i, \quad B=b-a,\quad C=|b-a|$$ In a time-crunch situation, I would be happy to submit my answer as
$$\int_0^1 |A+tB|^2 Cdt =|b-a|\left(|a-i|^2 + \Re\big((a-i)\overline{(b-a)}\big) + \frac13|b-a|^2\right).$$