Taylor Series: Can't quite work it out

AI Thread Summary
The discussion revolves around the application of Taylor series to derive a formula involving complex numbers and their conjugates. A user struggles to manipulate the equation z* = -iR + R^2/(z-iR) into the form suggested by their lecturer, which includes a Taylor series expansion. Participants suggest factoring and simplifying the expression, with one providing a corrected approach that involves manipulating the terms to clarify the placement of the imaginary unit 'i'. The conversation highlights the challenges of working with complex expansions and the importance of careful algebraic manipulation. Overall, the thread emphasizes the collaborative effort to resolve confusion around Taylor series applications in complex analysis.
wizard147
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Hi Guys,

Looking at some notes i have on conformal mapping and I have the following

where z is complex and z* denotes its conjugate, R is a real number


z* = -iR + R^2/(z-iR)

and my lecturer says that using the taylor series we get,

z* = -iR + iR(1+ z/iR + ...)

I've been trying for ages but I can't get this, I'm probably doing something stupid.

Anybody point me in the right direction? I'm getting confused with all these expansions!

I'm doing the following:

z* = -iR + R^2/z(1-iR/z)

and using the formula (1-x)^-1 = 1 + x^2 + x^3 (reference to wiki)

but it's not quite working! Hope you can help

C
 
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hi wizard147! :smile:

(try using the X2 button just above the Reply box :wink:)
wizard147 said:
z* = -iR + R2/(z-iR)

and my lecturer says that using the taylor series we get,

z* = -iR + iR(1+ z/iR + ...)

you just have to fiddle around with it a little :wink:

z* = -iR + R2/(z-iR)

= -iR + R/(z/R - i)

= -iR + iR/(z/iR - 1) …​

hmm, there's a sign wrong somewhere :redface:
 
Hi Tim,

Thanks, I think when you factored out your i, and then multiplied top and bottom by i you would get

-iR + iR/(1-z/iR)

I could be wrong though! lol
 
yup! …

i think i got confused about whether the last "i" was on the top or the bottom of iR/z/iR ! :biggrin:

thanks! :smile:
 
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