Exercise · ZW-COMPLEX-LIMIT-009 · ★★★ Advanced

If \(|f(z)-(2+i)|\le 3|z|\) for all \(z\), what follows about \(\lim_{z\to 0}f(z)\)?

Choose one option.

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  • C. it equals \(2+i\)

Explanation

Since \(3|z|\to 0\), the modulus \(|f(z)-(2+i)|\) tends to \(0\) (squeeze theorem), i.e. \(f(z)\to 2+i\).

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Review the concept: Complex limit

The limit \(L\) of a function \(f\) at \(z_0\) is the complex number that \(f(z)\) gets as close to as desired (\(\varepsilon\)) when \(z\) comes within distance \(\delta\) of \(z_0\), from any direction.

\[\lim_{z\to z_0}f(z)=L\iff\forall\varepsilon>0\ \exists\delta>0:\ 0<|z-z_0|<\delta\Rightarrow|f(z)-L|<\varepsilon\]

More practice problems: complex limit

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