Question

Which of the following is true about f(z)=\(\frac{z^2+(iz)^2}{z^2}\)?

(a) Continuous and differentiable

(b) Continuous but not differentiable

(c) Neither continuous nor differentiable

(d) Differentiable but not continuous

The question was asked in final exam.

My question comes from Differentiability topic in portion Complex Function Theory of Engineering Mathematics

Answer 1

The correct option is (a) Continuous and differentiable

To elaborate: z=x+iy

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

f(x, y)=\(\frac{(x+iy)^2+(ix+i^2y)^2}{(x+iy)^2}\)

f(x, y)=\(\frac{(x+iy)^2+(ix-y)^2}{(x+iy)^2}\)

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

\(\lim_{y \to 0}\frac{(iy)^2+(-y)^2}{(iy)^2}\)

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

=\(\frac{-1+1}{-1}\)

=0

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

\(\lim_{x \to 0}\frac{(x)^2+(ix)^2}{(x)^2}\)

\(\lim_{y \to 0}\frac{(x)^2-(x)^2}{(x)^2}\)

=\(\frac{1-1}{1}\)

=0

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

f(z)=\(\frac{z^2-z^2}{z^2}\)=0

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

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

=0

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