I am a beginner at calculating Cauchy Integrals but this one didn't look familiar to examples I have found. I think it is a linear integral.
Could anyone give ideas how to solve it? Thank you in advance.
$$I = \int_{0}^{4+2i} \!\!\!z^* \,\mathrm{d}z = ?\qquad\qquad (z^* = x-iy)$$
Since $z^*$ is not holomorphic you need to know which curve to integrate along, not just the start and end points. Once you know the curve, find a parametrization. If we assume that the curve is a straight line from $0$ to $4+2i$, one possible choice of parametrization is $z(t) = t(4+2i)$, $0 \le t \le 1$. By definition, the value of the integral is then $$ \int_0^1 z^*(t) z'(t)\,dt = \int_0^1 t(4-2i)\cdot(4+2i)\,dt = 20\int_0^1 t\,dt = 10. $$ Note that another choice of curve probably gives a different value for the integral.