Hi, this is a question about Laurent series for complex analysis. I have done the partial fractions part, but I do not know how to do the next part. I'm completely lost for that !
#Laurent Series
blissful dagger
silver sigilBOT
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I'm not sure I'll be a lot of help, but I will propose what I believe would be a solution and hopefully it helps
gleaming bronzeBOT
shsgd
blissful dagger
Hiya I still don't get it
plain sigil
yeah I didnt mean to send that message yet, just editingg it now
shit, just lost that work. give me a min
plain sigil
An expansion about $z_0 = 1$ encapsulates one of the singularities already, which is helpful. So we can define an area of expansion to be $0 < |z-1| < 9$
Recall the expansion
$\frac{1}{1-w} = \sum_{k=0}^{\infty} {w^k}$ for $|w| < 1$
We want to use this expansion. To do so, we manipulate $f(z)$ to look like $\frac{1}{1-w}$
Since we are expanding about $z_0 = 1$, we can ignore the $z-1$ singularity and focus on $z+8$ singularity. So we have that,
$\frac{1}{z+8} = \frac{1}{(z-1)+8+1} = \frac{1}{(z-1)+9}$
Factor out 1/9
$\frac{1}{9} \cdot \frac{1}{1-(-\frac{z-1}{9})}$
Using the expansion from before, we have
$\begin{aligned}
\frac{1}{9} \cdot \sum_{k=0}^{\infty}{ ( -\frac{z-1}{9} )^k } &= \frac{1}{9} \cdot \sum_{k=0}^{\infty}{ \frac{(-1)^k}{9^k} \cdot (z-1)^k } \
&= \sum_{k=0}^{\infty}{ \frac{(-1)^k}{9^{k+1}} \cdot (z-1)^k }
\end{aligned}$
Now to add $z-1$ back in, we have
$f(z) = \frac{1}{z-1} \cdot \sum_{k=0}^{\infty}{ \frac{(-1)^k}{9^{k+1}} \cdot (z-1)^k } = \sum_{k=0}^{\infty}{ \frac{(-1)^k}{9^{k+1}} \cdot (z-1)^{k-1} }$
gleaming bronzeBOT
shsgd
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@blissful dagger hope this helps a bit
plain sigil
Because we create a disk centred at |z-1| with radius 9
imagine you have a tiny circle that surrounds the point z = 1 and grows until it reaches the next singularity
the result will be a circle such that all points strictly inside (none on the borders or at the centre) will satisfy 0 < |z-1| < 9
plain sigil
But how do we know it is 0<|z-1|<9 ?
It's also possible you may need to evaluate the region |z-1| > 9, but im not certain
blissful dagger
I don't get the next part
plain sigil
which part?
blissful dagger
Why do we factor out 1/9 ?
plain sigil
because we are trying to get 1 / z+8 to look like 1 / 1 - w
blissful dagger
Ohh I get it now
plain sigil
I did skip some algebraic manipulations, hopefully you can still see what I did there
then of course -(z-1/9) is our w
I forgot to state that the w is valid for the region of expansion
plain sigil
the last line?
blissful dagger
Yea
plain sigil
By the power rule $x^a / x^b = x^{a-b}$
gleaming bronzeBOT
shsgd
gleaming bronzeBOT
shsgd
blissful dagger
Idk how to make it to 1/1-r
plain sigil
what do you mean
blissful dagger
1/z-1 to 1/1-r
plain sigil
Since we are expanding about $z_0 = 1$, we know our expansion will have the term $(z-1)^k$ in it. So we delibrately ignore $\frac{1}{z-1}$ because we know we can pull it into the sum later
gleaming bronzeBOT
shsgd
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yeah, which is how we combine it with (z-1)^k to get (z-1)^k-1
plain sigil
Laurent series has a general form $\sum_{k=0}^{\infty} a_n \cdot (z-z_0)^k +\sum_{k=0}^{\infty} b_n \cdot (z-z_0)^{-k}$ right?
gleaming bronzeBOT
shsgd
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where $z_0$ is the point you are expanding around
gleaming bronzeBOT
shsgd
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in your case, $z_0 =1$ so you know you have $(z-1)^k and (z-1)^{-k}$ in the series right?
gleaming bronzeBOT
shsgd
blissful dagger
Yea
plain sigil
Now, since you know these terms are in the series expansion, you know you can pull in another $(z-1)^k$ term into the series
gleaming bronzeBOT
shsgd
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then you can pull in the 1/(z-1) from f(z)
so you don't include it in the 1/1-r expansion, because you pull it in after
blissful dagger
So now we are looking at that part (the principal part )?
plain sigil
the expansion of 1/(z+8) would correspond to the analytic part and 1/(z-1) would correspond to the principal part
blissful dagger
Oh yeah
I don't know how to do the principal part
1/(z-1) pull out
Sorry for being so dumb
plain sigil
complex analysis isn't exactly straight forward
i was completely lost when I first learnt this
blissful dagger
Everyone in my class hates the Laurent series so much
The delta-epilson thingy too
plain sigil
i don't remember going over delta-epsilon proofs in complex analysis, only in real analysis
anyway, to summarise:
we split $f(z)$ into $\frac{1}{z-1} \cdot \frac{1}{z+8}$
blissful dagger
We did the continuity at the beginning 😭
gleaming bronzeBOT
shsgd
plain sigil
then expand $\frac{1}{z+8}$ using the $\frac{1}{1-w}$ expansion
gleaming bronzeBOT
shsgd
plain sigil
We did the continuity at the beginning 😭
I was awful at those, probably still am
blissful dagger
I thought u needed to use the partial fraction because my lecturer used the partial fractions
But everyone was lost 😭
blissful dagger
**shsgd**
But we can't do that in the series! We can't do sum*sum, but we can sum+sum
plain sigil
I don't remember the partial fractions route, would have to look over a sample solution to jog my memory
blissful dagger
I know we are very close
plain sigil
But we can't do that in the series! We can't do sum*sum, but we can sum+sum
it becomes $\frac{1}{z-1} \cdot \sum$
gleaming bronzeBOT
shsgd
plain sigil
and then it gets pulled into the series
plain sigil
yes except 9^n+1
blissful dagger
Ok typo
plain sigil
then when you pull the 1/z-1 in, the z-1 term in the summand becomes (z-1)^n-1
blissful dagger
plain sigil
yes exactly
blissful dagger
I thought there were two parts
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note that this includes both analytic and principal part
blissful dagger
But this one has only one part
plain sigil
when $n=0$ you have $\frac{1}{9} \cdot (z-1)^{-1}$
gleaming bronzeBOT
shsgd
blissful dagger
**shsgd**
But why is this the principal part ?
Because it's the negative power ?
plain sigil
yes
plain sigil
it gets like that when it comes to topics like complex analysis
blissful dagger
How do I close this forum when I'm done?
plain sigil
.solved
plain sigil
is this not the help channel?
blissful dagger
This is the help forum
plain sigil
no clue what channel you are on about then
blissful dagger
Ohh thank you so much anyway, I am appreciate your help
plain sigil
👍 good luck with the exam
blissful dagger
.solved