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I need some to help me verify my answer is correct: Let $A \subseteq \mathbb{C}$ and let $f:A\to \mathbb{C}$ be continuous at $z_0 \in A$. Let $f(z_0)\not=0$.

Prove that there exists $\delta >0$ such that $f(z)\not=0$ whenever $z\in B(z_0;\delta)$.

My Attempt: Let $f(z_0)=a$. Let $\epsilon = \frac{|a|}{2}$. Then $\epsilon > 0$. Since $f$ is continuous there exists $\delta > 0$ such that for all $z$, $0 < |z - z_0| < \delta$ implies $|f(z) - a| < \epsilon$. By the triangle inequality, if $0 < |z - z_0| < \delta$, then $$|f(z)| \ge |a| - |f(z) - a| > |a| - \epsilon = \frac{|a|}{2}.$$ Thus $f(z)> 0$ in the neighborhood of $z_0$.

Note: $$|a| = |a - f(z) + f(z)| \le |a - f(z)| + |f(z)| = |f(z) - a| + |f(z)|$$ so $|f(z)| \ge |a| - |f(z) - a| $.

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    You need to change your $f(z)\geq 0$ to $|f(z)| > 0$ in the conclusion. Apart from that it's correct. Well done. – 2017-02-11

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