I need some help and correct my knowledge, please.
Let $f(z)=(e^{z}-1)/(1+z+z^{2})$. Determine the largest domain $\Omega$ in $\mathbb{C}$ such that $f$ is holomorphic in $\Omega$.
Since $1+z+z^{2}=0\Longleftrightarrow z=\frac{-1\pm\sqrt{3}i}{2}$, then $f$ is holomorphic in $\Omega=\mathbb{C}\setminus \left \{ \frac{-1\pm\sqrt{3}i}{2} \right \}$.
Determine the radius of convergence of the Taylor series of $f(z)$ with center $a=1/2$.
How do I determine $\rho=\inf\lbrace|a-z|:z\in \Omega\rbrace$? Otherwise it would take time to determine a Taylor series in order to find its radius of convergence.
Show that Taylor series of $f(z)$ with center $a = 0$ is given by $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$, where $a_{0}=0$, $a_{1}=-\frac{1}{2}$ and $a_{k}=(1-k)/k! + a_{k-3}$ for $k\geq 3$. Use them to determine $f^{(6)}(0)$.
I will use $a_{k}=f^{(k)}(0)/k!$. First we see that $$ f'(z)=\frac{e^{z}(1+z+z^{2})-(e^{z}-1)(1+2z)}{(1+z+z^{2})^{2}}=\frac{f(z)(z^{2}-z)+1}{1+z+z^{2}} $$ and \begin{align*} f''(z)&=\frac{(f'(z)(z^{2}-z)+f(z)(2z-1))(1+z+z^{2})-(f(z)(z^{2}-z)+1)(1+2z)}{(1+z+z^{2})^{2}}\\ &=f(z)\frac{(2z-1)}{1+z+z^{2}}-f'(z)\frac{(z^{2} - 3z - 1)}{1+z+z^{2}}\end{align*} so determining a general expression of $f^{(k)}(z)$ will be awfully difficult.
If $f$ is holomorphic in $\Omega$, and $D(a,r)\subset \Omega$ (where $D(a,r)$ is the disc centered at $a$ with radius $r$), then the radius of convergence of the power series expansion of $f$ around $a$ it at least $r$. If $f$ can not be extended holomorphically in any larger disc, then the radius of convergence of the power series expansion of $f$ around $a$ it EXACTLY $r$.
In our case, $$ D\left(\frac{1}{2},\frac{\sqrt{7}}{2}\right)\subset \Omega=\mathbb C\smallsetminus\big\{\frac{-1\pm i\sqrt{3}}{2}\big\}. $$ But $$ \frac{-1\pm i\sqrt{3}}{2}\in \overline{D}\left(\frac{1}{2},\frac{\sqrt{7}}{2}\right), $$ as $$ \left|\frac{1}{2}-\frac{-1\pm i\sqrt{3}}{2}\right|=\left|1+\frac{\mp i\sqrt{3}}{2}\right| =\frac{\sqrt{7}}{2}. $$ and hence $f$ can not be extended holomorphically in any larger disc.
Thus the radius of convergence of the power series expansion of $f$ around $\dfrac{1}{2}$ is exactly $\dfrac{\sqrt{7}}{2}$.