2

Suppose f and g are analytic on a domain G. If f and g are non-constant, then for any b in G, there exists a punctured disk D'(b,R) of radius R>0 such that f(z)g(z) is not equal to g(z)-f(z) + 1 for all z in D'(b,R) IS this sentence incorrect?

Thank you in advance

aD1234567890
  • 133

1 Answers1

1

The statement is correct. It becomes more transparent if you rewrite $f(z)g(z)\ne g(z)-f(z)+1$ as $$f(z)\,(g(z)+1) \ne g(z)+1 \tag{1}$$ The two ways for equality to hold in (1) is that $f=1$ or $g=-1$. Since the sets $\{z:f(z)=1\}$ and $\{z:g(z)=-1\}$ are discrete, there is a punctured neighborhood of $b$ that does not meet their union.