Proving $f(z)$ is continuous on whole complex plane

Question

Question: define a function,

$$f(z)=\begin{cases}\frac{z \text{ Re(z)}}{|z|}&\text{ if $z≠0$}\\0, &\text{ if $z=0$}\end{cases}$$

Then prove or disprove $f(z)$ is continuous in entire complex plane.

We can see by definition of $f$ it is not constant function, so we can't use Liouville's theorem to prove $f$ is constant :-(

So I think we have to go by definition. I know, the function $f(z)$ is said to be continuous at $z=z_0$ if for given ε>0 we can find δ>0 such that,

$|f(z)-f(z_0)|<ε$ whenever $|z-z_0|<δ$

Now here to prove the continuity of $f(z)$ on whole complex plane, we just need to check continuity of $f(z)$ at $z=0$ only(?is am I correct)

Now,

$|f(z)-f(0)|=|\frac{z\text{ Re(z)}}{|z|}|$

$≤\frac{|z\text{ Re(z)}|}{|z|}$

$≤|Re(z)|$

But I end up with no conclusion? :-( is $f$ is continuous? I tried it, but didn't able to go further, Please help me...

Answer 1

No, it is not enough to prove that it is continuous at $0$. But outside $0$ it is continuous, becaue you can obtain $f$ from continuous functions using multiplication and division.

At $0$ it is continuous, because it follows from what you did that$$|z|<\varepsilon\implies\bigl|f(z)\bigr|<\varepsilon,$$since$\bigl|f(z)\bigr|\leqslant|z|$.

Answer 2

So $|f(z)-f(0)|\leq|\text{Re}(z)|\leq\sqrt{|\text{Re}(z)|^{2}+|\text{Im}(z)|^{2}}=|z|$, you may conclude now.

For other points $w\ne 0$ and consider $z$ closed to $w$, then \begin{align*} &\left|\dfrac{z\text{Re}(z)}{|z|}-\dfrac{w\text{Re}(w)}{|w|}\right|\\ &=\left|\dfrac{z|w|\text{Re}(z)-z|z|\text{Re}(z)+z|z|\text{Re}(z)-w|z|\text{Re}(w)}{|z||w|}\right|\\ &\leq\dfrac{|z||\text{Re}(z)|\big||w|-|z|\big|}{|z||w|}+\dfrac{|z||z\text{Re}(z)-w\text{Re}(w)|}{|z||w|}\\ &=\dfrac{|\text{Re}(z)||w-z|}{|w|}+\dfrac{|z\text{Re}(z)-z\text{Re}(w)|+|z\text{Re}(w)-w\text{Re}(w)|}{|w|}\\ &\leq\dfrac{|\text{Re}(z)||w-z|}{|w|}+\dfrac{|z||\text{Re}(z)-\text{Re}(w)|}{|w|}+\dfrac{|\text{Re}(w)||z-w|}{|w|}\\ &\leq\dfrac{|\text{Re}(z)||w-z|}{|w|}+\dfrac{|z||z-w|}{|w|}+\dfrac{|\text{Re}(w)||z-w|}{|w|}. \end{align*} For $\epsilon>0$, choose $\delta=\min\{|w|(1+|w|)^{-1}\epsilon/3,|w|/2,1\}$, for all $z$ with $|z-w|<\delta$, we have in particular that $|z-w|<1$, so $|z|<1+|w|$ by triangle inequality, and that $|\text{Re}(z)|\leq|z|<1+|w|$, we also have $|z-w|<|w|/2$, so $|z|>|w|-|w|/2=|w|/2>0$, and hence \begin{align*} &\left|\dfrac{z\text{Re}(z)}{|z|}-\dfrac{w\text{Re}(w)}{|w|}\right|\\ &\leq\dfrac{3(1+|w|)}{|w|}|w-z|+\dfrac{1+|w|}{|w|}|z-w|+\dfrac{|w|}{|w|}|z-w|\\ &<\dfrac{3(1+|w|)}{|w|}|w-z|\\ &<\dfrac{3(1+|w|)}{|w|}\delta\\ &<\dfrac{3(1+|w|)}{|w|}\dfrac{|w|}{1+|w|}\dfrac{\epsilon}{3}\\ &=\epsilon. \end{align*}

Answer 3

$|Re(z)|$ is continuous, so there exists some small $\delta$ such that $|Re(z)|$ is small if $|z|<\delta$.