• Support PF! Buy your school textbooks, materials and every day products Here!

Compute the taylor's expansion (series)

  • Thread starter MohdPrince
  • Start date
  • #1
[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



i need answer as soon as possible please!
 

Answers and Replies

  • #2
malawi_glenn
Science Advisor
Homework Helper
4,786
22
we don't give answer here, read and follow the forum rules.
 
  • #3
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

and i dont know what to do after that, please help me!
thanks...
 
  • #4
Vid
401
0
Start from the definition of a Taylor series.
 
  • #5
malawi_glenn
Science Advisor
Homework Helper
4,786
22
this is complex analysis right?

Are you supposed do evaluate taylor series for this f(z) around z_0 = 1 ?
 
  • #6
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?
 
  • #7
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
 
  • #8
Vid
401
0
Divide z+2 into the top. Break apart the resulting fraction and simplify.
 
  • #9
can you please show me the steps of dividing by z+2 & the breaking apart of the fraction?

thanks
 
  • #10
Vid
401
0
No..
 
  • #11
malawi_glenn
Science Advisor
Homework Helper
4,786
22
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.
 
  • #12
ok thanks very much, i'll try solving it again
 
  • #13
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! + ....
 

Related Threads for: Compute the taylor's expansion (series)

Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
924
  • Last Post
Replies
12
Views
7K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
960
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
3K
Top