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Dark Soul's question at Yahoo! Answers (Laurent expansion)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Expand the function in a Laurent series with center Z=Zo and determine the precise region of convergence.

(z^3 - (2 iota z^2) )
---------------------------- , Zo= iota
(z-iota)^2
Here is a link to the question:

Find Laurent series, please help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Dark Soul,

Denoting $u=z-z_0$ we get $$\frac{z^3-2z_0^2}{(z-z_0)^2}=\frac{(u+z_0)^3-2z_0^2}{u^2}=\frac{u^3+3u^2z_0+3uz_0^2+3z_0^3-2z_0^2}{u^2}\\=\frac{3z_0^3-2z_0^2}{u^2}+\frac{3z_0^2}{u}+3z_0+u=\frac{3z_0^3-2z_0^2}{(z-z_0)^2}+\frac{3z_0^2}{z-z_0}+3z_0+(z-z_0)$$ The Laurent expasion corresponds to a finite sum, so the expansion is valid for all $z\neq z_0$ (i.e. in $0<\left|z-z_0\right|<+\infty$).
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Here is the question:



Here is a link to the question:

Find Laurent series, please help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
\(\displaystyle \displaystyle \begin{align*} f(z) &= \frac{z^3 - 2\iota \, z^2 }{ \left( z - \iota \right) ^2} \\ &= -\left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left( \frac{1}{z - \iota} \right) \\ &= - \left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \frac{1}{z} \left( \frac{1}{ 1 - \frac{\iota}{z} } \right) \right] \\ &= - \left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \frac{1}{z} \sum_{n = 0}^{\infty} { \left( \frac{\iota}{z} \right) ^n } \right] \textrm{ provided } \left| \frac{\iota}{z} \right| < 1 \\ &= -\left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \sum_{n = 0}^{\infty} \left( \frac{ \iota ^n}{ z^{n + 1} } \right) \right] \textrm{ provided } |z| > 1 \\ &= - \left( z^3 - 2\iota \, z^2 \right) \sum_{n = 0}^{\infty} \left[ \frac{ -\iota ^n \left( n + 1 \right) }{ z^{n+2} } \right] \\ &= \left( z - 2\iota \right) \sum_{n = 0}^{\infty} \left[ \frac{ \iota^n \left( n + 1 \right) }{ z^n } \right] \end{align*}\)
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
\(\displaystyle \displaystyle \begin{align*} f(z) &= \frac{z^3 - 2\iota \, z^2 }{ \left( z - \iota \right) ^2} \\ &= -\left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left( \frac{1}{z - \iota} \right) \\ &= - \left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \frac{1}{z} \left( \frac{1}{ 1 - \frac{\iota}{z} } \right) \right] \\ &= - \left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \frac{1}{z} \sum_{n = 0}^{\infty} { \left( \frac{\iota}{z} \right) ^n } \right] \textrm{ provided } \left| \frac{\iota}{z} \right| < 1 \\ &= -\left( z^3 - 2\iota \, z^2 \right) \frac{d}{dz} \left[ \sum_{n = 0}^{\infty} \left( \frac{ \iota ^n}{ z^{n + 1} } \right) \right] \textrm{ provided } |z| > 1 \\ &= - \left( z^3 - 2\iota \, z^2 \right) \sum_{n = 0}^{\infty} \left[ \frac{ -\iota ^n \left( n + 1 \right) }{ z^{n+2} } \right] \\ &= \left( z - 2\iota \right) \sum_{n = 0}^{\infty} \left[ \frac{ \iota^n \left( n + 1 \right) }{ z^n } \right] \end{align*}\)
Notice that the problem says: centered at $z=z_0.$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403