# Compute the taylor's expansion (series)

[SOLVED] compute the taylor's expansion (series)

## Homework Statement

compute the taylor's expansion (series) for:

f(z)= [z^4 + (2-3i)*z^3 - 6i*z^2 + 2]/[z(z+2)]

where Zo (Z node or Z(0)) = 1

malawi_glenn
Homework Helper

i tried replacing each z wth x+iy but it went into a huge binomial equation i couldnt solve, here is part of my work:
z^4 = (x+iy)^4 = x^4 + (x^3 + iy) + (x^2 - y^2) + (x - iy^3) + y^4

z^4 + (2-3i)*z^3 = x^4 + (3-3i)*(x^3 + x^2 + x - y^2) + (3 + 3i)*(y - y^3) + y^4

thanks...

Start from the definition of a Taylor series.

malawi_glenn
Homework Helper
this is complex analysis right?

Are you supposed do evaluate taylor series for this f(z) around z_0 = 1 ?

this is complex analysis right?

Are you supposed do evaluate taylor series for this f(z) around z_0 = 1 ?

yes, can you help in that?

Start from the definition of a Taylor series.

the problem is that this equation will not be easy to defferenciate it has to be simplified first
or if you could do it as it is please tell me how

thanks

Divide z+2 into the top. Break apart the resulting fraction and simplify.

can you please show me the steps of dividing by z+2 & the breaking apart of the fraction?

thanks

No..

malawi_glenn
Homework Helper
can you please show me the steps of dividing by z+2 & the breaking apart of the fraction?

thanks

We dont provide that kind of help here. You show that you can try first, then we help you.

Takning the derivatives of [z^4 + (2-3i)*z^3 - 6i*z^2 + 2]/[z(z+2)] is straightforward if you know your derivation rules, which are the same as in real analysis.

ok thanks very much, i'll try solving it again

thanks for all your help, i've finally done it, and here is my final answer:

f(z)= (5/3 - 3i) + (10/9 - 3i) (z - 1)/1! + (50/27) [(z-1)^2]/2! + (4/27) [(z - 1)^3]/3! + ....