Is $\frac{\Re (z)}{z}$ continuous at $z=0$?

Question

How would I show that $\frac{\Re (z)}{z}$ is continuous at $z=0$? I know that the real value of a complex number equals $\frac{z+\bar z}2$, but I'm not sure where to go when I have written out the problem once I express $\Re (z)$ in this form.

Answer 1

Suppose $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , so

$$\frac{\text{Re}\,z}z=\frac x{x+iy}=\frac{x^2-ixy}{x^2+y^2}$$

If you now choose to approach the origin via the line $\;y=mx\;$ , you get:

$$\frac{x^2(1-im)}{x^2(1+m^2)}\xrightarrow[z\to 0\iff (x,y)\to (0,0)]{}\frac{1-im}{1+m^2}$$

so the limit depends on $\;m\;$ and can't thus exist...

Answer 2

That function is not continuous at $z=0$ since along the real line the function is always $1$ but along the imaginary line the function is $0$.