Why does a different parametrization give a different answer?

Question

I was doing this complex integration question:

$$\int_{c}f(z)dz$$ where $f(z) = z^{2}-3$ and the curve is the union of segments $[0,i]$ and $[i,1+2i]$. I tried to parametrize the curve in two different ways and see what I got.

First parametrization:

$C = it,\;0\le t \leq1$ and $C = t-1+it,\;1\le t \leq2$

$$\int_{c}f(z)dz=\int_{0}^{1}f(it)(i)dt + \int_{1}^{2}f(t-1+it)(i+1)dt$$

when I evaluate this, I end up getting

$$\frac{-20}{3}(1+i)$$

The second parmaterization I picked was

$C = it,\;0\le t \leq1$ and $C = (1+i)t +i,\;0\le t \leq1$

This time when I compute the integral, I get

$$\int_{c}f(z)dz=\int_{0}^{1}f(it)(i)dt + \int_{0}^{1}f(t+it+i)(i+1)dt$$

to be equal to $-(\frac{20}{3} + \frac{61i}{12})$

The answer to the first integral was from an answer key I found online and the second integral I evaluated using wolfram alpha. I thought that it shouldn't matter which way you parameterize it. I was wondering whether someone can explain why I'm getting different answers?

Answer 1

Since the left summands are equal in both expressions, I only worried about the right ones. I entered both into wolfram alpha and let it compute the difference, for me it returned zero. My exact query:

integral ((it + t - 1)^2 - 3) * (i+1) dt from 1 to 2 - integral ((i*t + t + i)^2 - 3) * (i+1) dt from 0 to 1

I apologize for not adding this as a comment, my reputation does not allow that yet.