Determine whether each statement is true, proving it or giving a counterexample.
If \(f,g\) have a pole at \(a\), then \(f+g\) has a pole at \(a\).
If \(f,g\) have a pole at \(a\), then \(fg\) has a pole at \(a\).
If \(f\) has an essential singularity at \(z_0\) and \(g\) has a pole at \(z_0\), then \(z_0\) is an essential singularity for \(f+g\).
If \(f\) has a pole of order \(N\) at \(z_0\) then \(f^2\) has a pole of order \(2N\) at \(z_0\).
Solution
The first statement is false. Take \[ f(z)={1\over z-a}, \qquad g(z)=-{1\over z-a}; \] then both have simple poles at \(a\), but \(f+g=0\).
The second statement is true. If \(f\) and \(g\) have pole orders \(m,n\), write \[ f(z)=(z-a)^{-m}u(z), \qquad g(z)=(z-a)^{-n}v(z), \] where \(u,v\) are holomorphic and nonzero at \(a\). Then \[ f(z)g(z)=(z-a)^{-(m+n)}u(z)v(z), \] and \(u(a)v(a)\neq0\), so \(fg\) has a pole of order \(m+n\).
The third statement is true. If \(g\) has a pole of order \(N\), its Laurent series has no terms below degree \(-N\). An essential singularity of \(f\) has infinitely many nonzero negative Laurent coefficients. Therefore the coefficients of \(f+g\) below degree \(-N\) are exactly those of \(f\), so infinitely many remain; hence \(f+g\) is essential at \(z_0\).
The fourth statement is true. Write \[ f(z)=(z-z_0)^{-N}u(z), \qquad u(z_0)\neq0. \] Then \[ f(z)^2=(z-z_0)^{-2N}u(z)^2, \] and \(u(z_0)^2\neq0\), so \(f^2\) has a pole of order \(2N\).