- #1

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**[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!**

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- Thread starter MohdPrince
- Start date

- #1

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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

- #2

malawi_glenn

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we don't give answer here, read and follow the forum rules.

- #3

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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

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Start from the definition of a Taylor series.

- #5

malawi_glenn

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

- #6

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

yes, can you help in that?

- #7

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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

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

- #9

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

thanks

thanks

- #10

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No..

- #11

malawi_glenn

Science Advisor

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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

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

- #13

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

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