Laurent series expansion of a function

Question

The question is : Show that $$e^{{c\over 2}(z-{i \over z})}={\sum_{n=-\infty}^\infty} a_nz^n$$

The question solved using Laurent's Series expansion where

$$a_n={1\over 2\pi i}\int_cf(z){dz \over z^{n+1}}$$ and $$b_n={1\over 2\pi i}\int_cf(z){dz \over z^{1-n}}$$

Then integrating and solving for $a_n$ i get

$$a_n={1\over 2\pi}\int_0^{2\pi}\cos(n\theta-c\sin\theta)d\theta$$

Now, we see that the function remains unaltered if $z$ is replaced by $-1\over z$

So far so good, now I don't understand the steps after this, it is given that :

so, $b_n=(-1)^na_n$ , It follows that

$$e^{{c\over 2}(z-{i \over z})}={\sum_{n=0}^\infty a_nz^n} + {\sum_{n=1}^\infty b_nz^{-n}} = {\sum_{n=-\infty}^\infty a_nz^n}.$$

Basically I don't understand how $b_n=(-1)^na_n$ and also the last part where the summation has been extended to $-\infty$

Answer 1

IMHO, there could a typo in the power of $(-1)$ for the expression for $b_n$. This is what I got: substitute $y=-1/z$ in the integral for $a_n$, this gives $a_n = \int_c f(-1/y) (-y)^{n+1} y^{-2} dy = (-1)^{n-1} \int_c f(-1/y) (1/y)^{1-n} dy = (-1)^{n-1} \int_c f(y) (1/y)^{1-n} dy = (-1)^{n-1} b_n$.

Therefore, $b_n = (-1)^{n-1} a_n$.

Maybe I made a sign error somewhere, in which case $b_n = (-1)^{n} a_n$.

Answer 2

For the second part, I will start from $b_n = (-1)^n a_n$. Then $\sum^{+\infty}_{n=-\infty} a_n z^n = (\sum^{-1}_{n=-\infty} + \sum^{+\infty}_{n=0}) a_n z^n $. In the first sum here, transform the (dummy) summation index to $m=-n$, then this sum becomes $\sum^{+1}_{m=+\infty} a_{-m} z^{-m}$. If you now look at the integral expressions for $a_n$ and $b_n$ you gave, you see that $a_{-n}$ equals $b_n$. Hence substituting in the first sum gives $\sum^1_{+\infty} b_m z^{-m}$, which is the same as $\sum^{+\infty}_{1} b_m z^{-m}$, i.e., the second sum in your expression.