The integral $$I_1 = \int_{C} \bar zdz=4\pi i ,$$ when $C$ is the right-hand half $z=2e^{iθ},\ (-\dfrac{π}{2}≤θ≤\dfrac{π}{2})$, of the circle $|z| =2$ from $z =-2i$ to $z = 2i$. And, one also can show the integral $$I_2 = \int_{C} \dfrac {1}{z}dz=\pi i ,$$ holds for the same $C$.
Is this argument true? : Since $z \bar z = |z|^2=4$ so $$4 \pi i = 4\int_{C} \dfrac {1}{z}dz = z \bar z \int_{C} \dfrac {1}{z}dz= \int_{C} z \bar z\dfrac {1}{z}dz = \int_{C} \bar zdz=4\pi i.$$
In other words, $z \bar z$ is constant so it can 'pass into' the integral but it can't cancel out $z$ in the denominator as now it behaves as a function. Am I right?
You're right that on the given circle $z\bar z$ is a constant, and as long as you make sure that you're only dealing with values strictly on the circle (i. e. neither off the circle, nor differentiating away from the circle), you can interchange it with that constant. You are actually cancelling out the $z$ in the denominator, so I'm not sure what you mean by "but it can't cancel out $z$ in the denominator as now it behaves as a function".