Let f=u+iv with u=(xy)/(x^2+y^2), v=x^2 for z≠ 0 and f(0)=0. Is it continuous at 0?

No, because u is not continuous at (0,0)

Exercise · ZW-COMPLEX-CONTINUITY-007 · ★★☆ Intermediate

Let \(f=u+iv\) with \(u=\dfrac{xy}{x^2+y^2}\), \(v=x^2\) for \(z\neq 0\) and \(f(0)=0\). Is it continuous at \(0\)?

Choose one option.

Show answer

D. No, because \(u\) is not continuous at \((0,0)\)

Explanation

\(f\) is continuous at \(z_0\) if and only if \(u\) and \(v\) are at \((x_0,y_0)\). On \(y=0\), \(u=0\), but on \(y=x\), \(u=1/2\): \(u\) has no limit at \((0,0)\).

Study Complex analysis in ZWMath →

Review the concept: Complex continuity

A function \(f\) is continuous at a point \(z_0\) of its domain if \(f(z)\) tends to the value \(f(z_0)\) as \(z\) tends to \(z_0\); it is continuous if it is so at every point of its domain.

\[\lim_{z\to z_0}f(z)=f(z_0)\]

Let \(f=u+iv\) with \(u=\dfrac{xy}{x^2+y^2}\), \(v=x^2\) for \(z\neq 0\) and \(f(0)=0\). Is it continuous at \(0\)?

Explanation

Review the concept: Complex continuity

More practice problems: complex continuity