# complex analytic functions

• May 13th 2010, 07:58 PM
sandy
complex analytic functions
hi to all can i please see the answer for this to check what i have been doing wrong thanks heaps.
(a) Where is the function
f(z) = z/(1+iz)^4 analytic?

(b) Find
f(-i).
• May 13th 2010, 08:09 PM
vincisonfire
a) Plug f(z) in the Cauchy-Riemann equations and see where they are satisfied
b) Replace z by -i
• May 13th 2010, 08:24 PM
Drexel28
Quote:

Originally Posted by sandy
hi to all can i please see the answer for this to check what i have been doing wrong thanks heaps.
(a) Where is the function
f(z) = z/(1+iz)^4 analytic?

(b) Find
f(-i).

Quote:

Originally Posted by vincisonfire
a) Plug f(z) in the Cauchy-Riemann equations and see where they are satisfied
b) Replace z by -i

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?
• May 16th 2010, 09:57 AM
sandy
Quote:

Originally Posted by vincisonfire
a) Plug f(z) in the Cauchy-Riemann equations and see where they are satisfied
b) Replace z by -i

can i please have a hint of how to break it up
do i substitute z by x+iy but then its to big
what do i do?
• May 16th 2010, 01:03 PM
vincisonfire
It is a bit messy.
It may be simpler in this form :
$\displaystyle f(z) = \frac{z}{(1+iz)^4} = \frac{r_1e^{i\theta_1}}{(r_2e^{i\theta_2})^{4}} = \frac{r_1e^{i\theta_1}}{r_2^4e^{4i\theta_2}}$
• May 17th 2010, 03:10 AM
sandy
Quote:

Originally Posted by vincisonfire
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??