Affine transformation of the curve for the complex line integral. I'm studying the efficient ways to calculate for the complex line integral.Though this question  looks like silly, I want to check my thought is right or not.
Let the complex function $f : \mathbb{C}\to \mathbb{C}$
Say the  curve $C_r$ on the complex plane. Define the function $\phi : C_r \to C'_r$ by $z\to w $ (Here the mapping $\phi$ is either rotation, the symmetry or scaling)
From this we get  $w= \phi (z) $  and $dw = \phi'(z) dz $
As the example, I took the $\phi$ that rotation as $\frac\pi 2$ counter-clockwise. (I.e. $\phi : z \to w(=iz)$)
Then Does the $\int_{C_r} f(z) dz = \int_{C_r'} f(\frac{w}{i})\frac{dw}{i}$ hold?
Plus If we generalize this thought, Could we say $\int_{C_r} f(z) dz = \int_{C_r'} f(\phi^{-1}(w)) \frac{dw}{\phi'(z)}$?
If my things incorrect, what is the exact form of that?
 A: Let $D \subset \Bbb C$ be an open set, and $\gamma: [a, b] \to D$ a (piecewise differentiable) path in $D$. For any continuous function $f: D \to \Bbb C$, the integral over $f$ along $\gamma$ is defined as
$$
 \int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, dt \, .
$$
Now let $\phi: D \to D' \subset \Bbb C$ be complex differentiable. Then $\Gamma = \phi \circ \gamma$ is a path in $D'$, and
$$
\int_\Gamma f(z) \, dz = \int_a^b f(\Gamma(t)) \Gamma'(t) \, dt
= \int_a^b f(\phi(\gamma(t)) \phi'(\gamma(t))\gamma'(t) \, dt
= \int_\gamma f(\phi(z))\phi'(z) \, dz\, .
$$
So we have the “substitution rule”
$$
 \boxed{\int_{\Gamma} f(z) \, dz =\int_\gamma f(\phi(z))\phi'(z) \, dz \, .}
$$
Roughly speaking: One can substitute $w = \phi(z)$ in a complex line integral, as long as one takes into a account that the new integral is along a different path.
Note that $\phi$ is not required to be an “affine transformation” here, just complex differentiable.
If $\phi$ happens to be invertible in $D'$ then $\gamma = \phi^{-1} \circ \Gamma$, and therefore
$$
 \int_{\gamma} f(w) \, dw =\int_\Gamma f(\phi^{-1}(w))\frac{1}{\phi'(\phi^{-1}(w))} \, dw \, .
$$
