The following functions are not defined at z=0 : 1^°…

The following functions are not defined at $z=0$: $\ 1^\circ\ \dfrac{\operatorname{Re}(z)}z;\quad2^\circ\ \dfrac z{|z|};\quad3^\circ\ \dfrac{z\operatorname{Re}(z)}{|z|^2}$. (a) Determine whether they have limits as $z\to0$. (b) Can they be extended to $z=0$ to become continuous?

Part $1^\circ$: $f(z)=\dfrac{\operatorname{Re}(z)}z$

Along the real axis ($z=x$): $f(x)=x/x=1$. Along the imaginary axis ($z=iy$): $f(iy)=0/(iy)=0$. Since $1\neq0$, $$ \lim_{z\to0}\frac{\operatorname{Re}(z)}z\ \text{does NOT exist} $$

Part $2^\circ$: $f(z)=\dfrac z{|z|}$

Writing $z=re^{i\theta}$, $f(z)=e^{i\theta}$ for every $r>0$ — independent of $r$ but fully dependent on the approach angle $\theta$. Along $\theta=0$: $f=1$; along $\theta=\pi$: $f=-1$; along $\theta=\pi/2$: $f=i$. Since these differ, $$ \lim_{z\to0}\frac z{|z|}\ \text{does NOT exist} $$

Part $3^\circ$: $f(z)=\dfrac{z\operatorname{Re}(z)}{|z|^2}$

With $z=x+iy$: $$ f(z)=\frac{(x+iy)x}{x^2+y^2}=\frac{x^2}{x^2+y^2}+i\,\frac{xy}{x^2+y^2} $$ Along the real axis ($y=0$): $f=1$. Along the imaginary axis ($x=0$): $f=0$. Along $y=x$: $f=\dfrac{x^2+ix^2}{2x^2}=\dfrac{1+i}2$. Three different values, so $$ \lim_{z\to0}\frac{z\operatorname{Re}(z)}{|z|^2}\ \text{does NOT exist} $$

Part (b) · Continuous extension

A function extends continuously to a point exactly when its limit exists there. Since none of the three limits above exist, none of the three functions can be made continuous at $z=0$ by any choice of $f(0)$ — each has an essential (non-removable) discontinuity at the origin.

Why

All three involve $\operatorname{Re}(z)$ or $|z|$, neither of which is holomorphic; mixing these real-valued, non-analytic quantities with $z$ or $1/z$ is exactly what produces the angle-dependent (path-dependent) behavior seen above.