Take the following complex integral: $$ \int_\gamma (z^2+2z)dz = -\frac{20}{3}+\frac{40}{3}i $$ where $\gamma$ is parameterised as $z(t)=1+t(1+2i)$ (so $dz=(1+2i)dt$) with $0\le t\le 1$.
For that particular integral, I solved it by doing the indeterminate integral in respect to $z$, then changing $z$ for $z(t)$. I was then asked to solve the variation: $$ \int_\gamma (z^2+2z)\|dz\| $$
The obvious way to solve it, for me, is to just replace right away and do the integral in respect to $t$. My question is, however: If I already have the first result, is there a way to get the result of the second integral without going through the hassle of solving it again?
i.e. is there a way to relate the first result with the second one just knowing what $dz$ and $\|dz\|$ are?
BONUS QUESTION: What does it mean to have $\|dz\|$ on an integral?
$\int_\gamma f(z)\,dz$ for a smooth curve $\gamma : [a,b] \to \Bbb C$ is defined by $$\int_\gamma f(z)\,dz :=\int_a^b f(\gamma(t))\gamma'(t)\,dt$$ $\int_\gamma f(z)\,\|dz\|$ is defined by $$\int_\gamma f(z)\,\|dz\| :=\int_a^b f(\gamma(t))|\gamma'(t)|\,dt$$
And there you see the problem: $\gamma'(t)$ can point in various directions as $t$ changes, but $|\gamma'(t)|$ only points in the direction of the positive reals. So while $\gamma'$ may twist around and in some places be adding positive and imaginary contributions, but elsewhere make negative contributions, with $|\gamma'|$ all contributions from $\gamma$ are positive and real. While $\int_\gamma f(z)\,dz$ and $\int_\gamma f(z)\,\|dz\|$ are intimately related, neither can be determined from the other.
Your trick for calculating the original integral works by Cauchy's theorem, and requires that the function $f$ be differentiable everywhere. If it isn't, then the integral value doesn't depend only on the endpoints, but also on the particular path between them being integrated. In that case, the value you get from an anti-derivative may match some paths, but not all of them, so it may or may not apply to the actual path of integration.
With $\int_\gamma f(z)\,|dz|$, this doesn't work at all. Cauchy's theorem doesn't apply to this integral. There is no guarantee that integrals along different paths will have the same value just because the endpoints are the same.
Likely the whole point of the exercise you are doing is to drive that point home to you. You will have to substitute in the curve and do the problem the hard way.