Prove that $\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1) }f(z)dz $

Question

Define $C(0,1)= \{z: \mid z\mid=1 \}$ $D(0,1)= \{z: \mid z\mid=1, z\neq 1+0i \}$

Prove that, $$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1) }f(z)dz $$ My try

$z=1+0i$ is of measure zero. So, $$\oint _{C(0,1)}f(z)dz= \oint _{ D(0,1) }f(z)dz $$

Answer 1

Of course. Changing the integrand on a set of measure zero does nothing to the integral, as you've essentially already noted. You could write the second integral as using the integrand $g(z) := f(z) 1_{z \neq 1}$ where $1_{z \neq 1}$ is $1$ if $z \neq 1$ and is $0$ if $z = 1$.

Answer 2

Since $$e^{i\theta}=1 \Rightarrow \theta=0,2\pi $$ $$\oint _{ D(0,1) }f(z)dz =\int_{[0,2\pi]- \{0,2\pi\} } ie^{i\theta}f(e^{i\theta})d\theta $$ Since $\{0,2\pi\} $ is a set of measure 0 $$\oint _{ D(0,1) }f(z)dz =\int_{0}^{2\pi} ie^{i\theta}f(e^{i\theta})d\theta $$ $$\oint _{ D(0,1) }f(z)dz =\oint _{ C(0,1) }f(z)dz$$