$\displaystyle \gamma \text{ is any curve joining } -1+i \text{ to }1.\\ \text{ Letting }\gamma(b)=1, \, \gamma(a)=-1+i \text{ and know that } (-1+i)^2 = -2i, $ $\displaystyle \begin{align*} \int_\gamma (z^3-6z^2 +4)dz &= \frac{\gamma^4(b)-\gamma^4(a)}{4}-6\frac{\gamma^3(b)-\gamma^3(a)}{3}+4(\gamma(b)-\gamma(a)) \\&= \frac{1^4- (-1+i)^4}{4}-6\frac{1^3- (-1+i)^3}{3}+4(1 - (-1+i)) \\ &= \frac{1^4- (-2i)^2}{4}-2(1^3- (-1+i)^2(-1+i))+4(2- i)) \\ &= \frac{1-(-4i)}{4} -2(1- (-2i)(-1+i)) +8 - 4i \\ &= \frac{1+ 4i}{4} -2(-1 -2i) +8 - 4i\\ &= \frac{41}{4} +i \end{align*} $

This question is taken from Fisher's "Complex Variables 2nd Edition" on page 74.

Is my computation correct? Let me know.

Thanks a lot.