What vincisonfire said is true, but surely you have proved that any rational function in \(\displaystyle \mathbb{C}\) is analytic everywhere it's denominator is non-zero, right?

It is a bit messy.
It may be simpler in this form :
\(\displaystyle f(z) = \frac{z}{(1+iz)^4} = \frac{r_1e^{i\theta_1}}{(r_2e^{i\theta_2})^{4}} = \frac{r_1e^{i\theta_1}}{r_2^4e^{4i\theta_2}} \)

It is a bit messy.
It may be simpler in this form :
\(\displaystyle f(z) = \frac{z}{(1+iz)^4} = \frac{r_1e^{i\theta_1}}{(r_2e^{i\theta_2})^{4}} = \frac{r_1e^{i\theta_1}}{r_2^4e^{4i\theta_2}} \)

thanks so much for your help i done it and found that cauchy riemann equations are satisfied every where so do we say f(z) is analytic everywhere on C??