Is $\frac{1}{z}$ continuous or not? Is $\frac{1}{z}$ continuous or not on $\mathbb C \setminus\{0\}$?
I was thinking 
$$\frac{1}{z}=\frac{1}{x+yi}=\frac{x-yi}{x^2+y^2}=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i$$
Since each part is continuous on domain, could we say $\frac{1}{z}$ is continuous?
Now from what I got below I know $\frac{1}{z}$ is continuous on its domain, so can we say since it's continuous so
$$\lim_{z\rightarrow z_0}\frac{1}{z}=\frac{1}{z_0}$$
More generous, can we have
$$\lim_{z\rightarrow z_0}\frac{1}{f(z)}=\frac{1}{\lim_{z\rightarrow z_0}f(z)}$$
 A: What you did is correct, but how do you know that each part is continuous?
You can prove the continuity directly. Fix $z_0\in\mathbb{C}\setminus\{0\}$. If $z\in\mathbb{C}\setminus\{0\}$, then$$\left|\frac1z-\frac1{z_0}\right|=\frac{|z-z_0|}{|z|.|z_0|}.$$If $|z-z_0|<\frac{|z_0|}2$, then $|z|=|z_0+z-z_0|\geqslant|z_0|-|z-z_0|>\frac{|z_0|}2$ and therefore$$\left|\frac1z-\frac1{z_0}\right|<\frac2{|z_0|^2}|z-z_0|.$$So, given $\varepsilon>0$, take $\delta=\min\left\{\frac{|z_0|^2}2\varepsilon,\frac{|z_0|}2\right\}$ and you'll have$$|z-z_0|<\delta\implies\left|\frac1z-\frac1{z_0}\right|<\varepsilon.$$
A: Yes, this clearly shows that the function $f(z) : \mathbb C \setminus (0,0) \to \mathbb C$ is continuous, since as you pointed out, one can see by letting the cartesian form of a complex number $z:=x + iy$ :
$$f(z) = \frac{1}{z} = f(x+iy) =\frac{1}{x+yi}=\frac{x-yi}{x^2+y^2}=\frac{x}{x^2+y^2}-\frac{y}{x^2+y^2}i $$
The real part $\Re\{f\}$ and the imaginary part $\Im\{f\}$ are continuous, which means that the complex function $f$ is continuous as well.
A: Addition, subtraction, and multiplication are always continuous, on the reals, complexes and many other domains. Division is also continuous, as long as the denominator does not vanish. (and where it does, the operation is not even defined, let alone continuous). So we can conclude that $1/z$ is continuous away from $z=0$ immediately, without any mention of its real and imaginary parts. 
We should prefer that anyway, to look at $1/z$ as a single operation rather than a combination of real and imaginary parts. What if someone were to ask you whether $1/z$ were complex differentiable? It is not enough then to look at real and complex parts separately...
A: Yes. If $\text{Re}f$ and $\text{Im}f$ are continuous, then $f$ is continuous.
