Question

Which of the following is true about f(z)=z^2?

(a) Continuous and differentiable

(b) Continuous but not differentiable

(c) Neither continuous nor differentiable

(d) Differentiable but not continuous

This question was addressed to me by my college professor while I was bunking the class.

The question is from Differentiability topic in section Complex Function Theory of Engineering Mathematics

Answer 1

Right option is (a) Continuous and differentiable

The best explanation: z=x+iy

In general the limits are discussed at origin, if nothing is specified.

f(x,y)=(x+iy)^2

Left limit: \(\lim_{x \to 0 \\ y \to 0}\)(x+iy)^2

\(\lim_{y \to 0}\)i^2y^2

=0

Right limit: \(\lim_{y \to 0 \\ x \to 0}\)(x+iy)^2

\(\lim_{x \to 0}\)x^2

=0

Both the limits are equal, therefore the function is continuous. To check differentiability,

f’(z)=\(\lim_{\delta z \to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\)should exist.

\(\lim_{\delta z \to 0}\frac{(z+\delta z)^2-(z)^2}{\delta z}\)

\(\lim_{\delta z \to 0}\frac{2z(\delta z)+(\delta z)^2}{\delta z}\)

=\(\lim_{\delta z \to 0}\)(2z+(δz))

=2z

Since f’(z) exists, the function is differentiable as well.