Uniform continuity of functions of a complex variable

Question

Let $f:\color{blue}D→\mathbb{C}$ be a complex function with $\color{blue}D = \{z \in \mathbb{C} \:/\:\frac12\leq|z|\leq1 \}$

$$f(z) = \frac{1+z^2}{3+z^3}$$

1st I have to find all the discontinuities, well that's easy there's only 3 which are the solutions of $3+z^3=0$

2nd I have to discuss the continuity so since it's a fraction it's continous everywhere except where it's not defined which means it's continous except at those 3 points

3rd I have to discuss the uniform continuity I know this definition : $$\forall\: \epsilon> 0\: \exists\:\delta>0\: \forall z_{\color{red}1},z_{\color{red}2},\:|z_{\color{red}1}-z_{\color{red}2}|<\delta\implies|f(z_{\color{red}1})-f(z_{\color{red}2})|< \epsilon $$

but I am stuck trying to figure out how to use the definition above to discuss the uniform continuity

if someone can push me in the right way or tell me how I can solve this that would be great.

Answer 1

Observe that the three roots of the denominator are outside $D$ and conclude that $f$ is continuous on $D$. Since $D$ is compact...

Answer 2

Hint: with $$f(z)=\frac{1+z^2}{3+z^3}=1-\frac{2}{3+z^3}$$ \begin{eqnarray} |f(z_{\color{red}1})-f(z_{\color{red}2})| &=& \Big|1-\frac{2}{3+z_1^3}-1+\frac{2}{3+z_2^3}\Big|\\ &=& 2\Big|\frac{z_2^3-z_1^3}{(3+z_1^3)(3+z_2^3)}\Big|\\ &=& 2\Big|\frac{(z_1-z_2)(z_1^2-z_1z_2+z_2^2)}{(3+z_1^3)(3+z_2^3)}\Big|\\ &\leq& 2|z_1-z_2|\frac{|z_1|^2+|z_1||z_2|+|z_2|^2}{(3+|z_1|^3)(3+|z_2|^3)}\\ &\leq& 2|z_1-z_2|\frac{3}{(3+(\frac12)^3)(3+(\frac12)^3)} \end{eqnarray}