Prove $f(z)=\frac{z}{z^4+1}$ is continuous within the unit circle in the complex plane centred at the origin

Question

The question is from Schaum's Outline of Complex Variables

Prove that $f(z)=\frac{z}{z^4+1}$ is continuous for all $|z|<1$ except four numbers in the complex plane. (Obviously the solutions to $z^4+1=0)$

I know that for $f$ to be continuous, $\forall\epsilon>0$, there exists a $\delta>0$ s.t, when $|z-z_0|<\delta$, $|f(z)-f(z_0)|<\epsilon$

$\to|\frac{z}{z^4+1}-\frac{z_0}{z_0^4+1}|$

$=|\frac{(z-z_0)(1-zz_0(z^2+zz_0+z_0^2))}{(z^4+1)(z_0^4+1)}|\le\frac{|z-z_0|(1+|zz_0|(|z|^2+|zz_0|+|z_0|^2))}{|(z^4+1)(z_0^4+1)|}<\frac{4|z-z_0|}{|(z^4+1)(z_0^4+1)|}$, first using the triangle inequality, then using the fact that both $|z|,|z_0|<1$

I am not quite sure how to manipulate the denominator of this expression to obtain the following result:

$\frac{4|z-z_0|}{|(z^4+1)(z_0^4+1)|}<k|z-z_0|$, where $k$ is a constant, probably dependent on $|z_0|$, so that I can complete the expsilon-delta proof.

Any help will be much appreciated. Thank you.

Answer 1

If $g,h\colon\Bbb C\setminus\{z\in\Bbb C\mid z^4\ne-1\}\longrightarrow\Bbb C$ are the functions defined by $g(z)=z$ and $h(z)=z^4+1$, then $g$ and $h$ are continuous and $h$ has no zeros. So, since $f=\frac gh$, $f$ is continuous.

Answer 2

$f$ is holomorphic except at those four points. You can use the quotient rule. Holomorphic $\implies$ continuous.