I make up my own questions and contours and solve the integration problem without knowing whether the answer is right or not. Please verify the below question for me:
$f(z) = z^2 + e^z$ and $\gamma(t)=t+(2t)i$. Calculate $\int_cf(z)dz$
Step1: Replace $z$ in $f(z)$ with $x+iy$. Expand the bracket. Bring like terms together.
Result: $f(z) = (x^2-y^2)+[2i(xy)+e^z]$
Step2: Calculate $\gamma\prime(t)$
Result: $1+2i$
Step3: Calculate $f(\gamma(t))$
Result: $-3t^2 + (4t^2)i + e^{t+2ti}$
Step4: Calculate $f(\gamma(t))\times\gamma\prime(t)$
Result: $-11t^2 - 2t^2i + e^{t+2ti}[1+2i]$
Step5: Calculate the integral
Result: $-\frac{11t^3}{3}-\frac{2it^3}{3}+e^{1+2ti}[1+2i]$
Step6: Evaluating from 0 to 1
Result $-\frac{11}{3}-\frac{2i}{3}+0$
You may have calculated the integral incorrectly. When you integrate $\int_0^1 e^{t+2ti}dt=\int_0^1 e^{(1+2i)t}dt=[\frac{1}{1+2i}e^{(1+2i)t}]_0^1=\frac{1}{1+2i}[e^{(1+2i)t}]_0^1$, and that saves you some work, because it will cancel out the other factor $1+2i$. I get that this leaves $e^{1+2i}-1$ plus what you get from the $z^2$.