= Solution
The <Liouville theorem> states that every bounded <entire function> is constant. Let $|f(z)|\le M$ and choose distinct $a,b$. For $R>\max(|a|,|b|)$, partial fractions and the <Cauchy integral formula> give
$$
I_R=\int_{|z|=R}\frac{f(z)}{(z-a)(z-b)}\,dz
=\frac{2\pi i}{a-b}[f(a)-f(b)],
$$
with the circle traversed anticlockwise. On that circle,
$$
|I_R|\le\frac{2\pi RM}{(R-|a|)(R-|b|)}\longrightarrow0.
$$
The displayed difference is independent of $R$, so $f(a)=f(b)$. Since $a,b$ were arbitrary, \b[$f$ is constant]. This proves the theorem with the requested two-pole contour rather than assuming it.
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