Question

expand f(z)=1/z-2 as a Taylor series in powers of z​

Answer 1

$\huge{\dag} \: {\underline{\boxed{\sf{\purple{ Answer }}}}}$

You may or may not know about the Binomial Theorem,[1] but if you don’t know about it, you should. It says that

For any non-negative integer, nn, (x+y)n=∑k=0n(nk)xn−kyk(x+y)n=∑k=0n(nk)xn−kyk.

But if you do know about this form of the theorem already, there’s a reasonable chance that you don’t know that there’s a version of it that works for other values of nn as well.

Recall that for integers 0≤k≤n0≤k≤n, (nk)=n!k!(n−k)!(nk)=n!k!(n−k)!. We can write this as (nk)=n(n−1)(n−2)…(n+1−k)k!(nk)=n(n−1)(n−2)…(n+1−k)k!.

Let’s extend this notation to mean this second version for all other values of nn as well. Using this notation, we can extend the Binomial Theorem to:

For any real number, nn, and for |x|>|y||x|>|y|, (x+y)n=∑k=0∞(nk)xn−kyk(x+y)n=∑k=0∞(nk)xn−kyk.

In the event that nn is a negative integer, with some algebra, we can rewrite the binomial theorem in an equivalent, simpler way as