We are supposed to integrate $z^2$ in the complex plane along an arc from $1$ to $i$ two ways, by parameterizing along an arc and by antiderivative.
This should be a straightforward question and I have done it before but somehow this time it is being obnoxious. Double-checked and worked out every trivial detail. Hopefully this is just a stupid error but I have obviously missed something. Help before I tear my hair out please.
By antiderivative, $$\int_{z=1}^i z^2 dz = \frac{z^3}{3}]_1^i = \frac{i^3}{3}- \frac{1^3}{3} = \frac{-1}{3} - \frac{i}{3}$$
Parameterizing, let $z = x+iy = (1-t) + it $ for $0 \leq t \leq 1$
When$ \ t = 0, z = 1$ and when $ \ t = 1, z = i$ as required.
$z^2 = ((1-t) + it)^2 = 1 - 2t + t^2 +2(1-t)(it) +(it)^2$
$= 1 - 2t + t^2 - t^2 +i(2t - 2t^2) = (1 - 2t) +i(2t - 2t^2) $
$$\int_{z=1}^i z^2 dz = \int_{t=0}^1[(1 - 2t) +i(2t - 2t^2)]dt$$
$$= \int_{t=0}^1(1 - 2t) dt + i\int_{t=0}^1 (2t - 2t^2)]dt$$
$$=(t -2t^2/2)]_0^1 + i(2t^2/2 - 2t^3/3)]_0^1$$
$$= (t -t^2)]_0^1 + i(t^2 - 2t^3/3)]_0^1$$
$$= 1 - 1^2 - 0 + 0^2 +i(1^2 - 2(1^3)/3 - 0^2 + 0^3)$$
$$= 0 - i(1/3) = 0 + i/3$$
So, why can't I make them match?
You forgot that $dz = (-1 + i)dt$. Also you have a sign error somewhere in your second integral; it should have come out to $i/3$.