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Problem about $\phi eq 1$ in $\int_0^{2\pi}e^{i\theta}\cdot\phi(e^{i\theta})d\theta$

If $\phi(z)$ be an analytic function, then $\phi eq 1$ in $\int_0^{2\pi}e^{i\theta}\cdot\phi(e^{i\theta})d\theta$

I can’t get the value of $\phi(e^{i\theta})$,
By expanding $\phi(e^{i\theta})$,
$$\phi(e^{i\theta})=\phi(e^{i\theta})+O(z-z_0)$$
So in the RHS, we get
$$\phi(e^{i\theta})=O(z-z_0)$$
Then,
$$e^{i\theta}\cdot\phi(e^{i\theta})=O(z-z_0)e^{i\theta}$$
And now
$$\int_0^{2\pi}e^{i\theta}\cdot\phi(e^{i\theta})d\theta=2\pi O(z-z_0)$$
As $|z-z_0|\leq 1$, and $\phi eq1$, we get
$$2\pi=O(z-z_0) eq 0$$
But I think the value must be zero. Where did I do wrong?

A:

If $\phi$ is analytic, then

\phi(e^{i