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.