Integrate $ f(z) $ counterclockwise around the unit circle. $$ f(z) = 1/(4z-3) $$
My solution
C(contour) : $ z(t) = \cos{t} + i\sin{t} = e^{it}, 0<t\leq 2\pi $
$$ \oint_C \frac{1}{4z-3} dz = \int_0^{2\pi} \frac{1}{4e^{it}-3} \frac{dz}{dt} dt = \int_0^{2\pi} \frac{1}{4e^{it}-3}ie^{it} $$ substitute $ u=e^{it}, du = ie^{it} dt $ $$ \oint_C \frac{1}{4z-3} dz = \int_1^1 \frac{1}{4u-3}du = 0 $$
But as you know, this answer is wrong. According to the Cauchy's integral formula, the answer must be $ \frac{i\pi}{2} $.
I cannot find an error in my way. I guess substitution $ u=e^{it} $ is wrong, but I don't know exactly. Why did I reach wrong result?