$$f(z) = \frac{e^z}{(z-2)^3}$$
I want to write Taylor series expansion for this function about $z=2$.
I know I can expand $e^z$ about $z=0$. So I proceed as follows:
Let $z-2=t$ so $z=2+t$
$$ f(t) = \frac{e^{2+t}}{t^3} $$
Now I can expand $e^{2+t}$ about t=0 as:
$ f(t) = \frac {1}{t^3} \left( 1 + \frac{(2+t)^1}{1!} + \frac{(2+t)^2}{2!}+.....\right) $
Now if I replace t by t=z-2, I end up with:
$ f(z) = \frac {1}{(z-2)^3} \left( 1 + \frac{(2+(z-2))^1}{1!} + \frac{(2+(z-2))^2}{2!}+.....\right) $
$ f(z) = \frac {1}{(z-2)^3} \left( 1 + \frac{(z)^1}{1!} + \frac{(z)^2}{2!}+.....\right) $
But this series is same as expansion of $f(z) = \frac{e^z}{(z-2)^3}$ about $z=0$
In the book they have done like this:
Please help me understand why the way I am writing the Taylor series is wrong and the way it is given in the book is right.