I need to expand $\frac{e^z}{(z-1)^2}$ about $z=1$. I tried substituting 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + ... for $e$ but it doesn't solve the purpose. How do I go about solving this?
Expand $\frac{e^z}{(z-1)^2}$ about $z=1$
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algebra-precalculus
exponential-function
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1Hint : $e^{z-1}=e^z\cdot \frac{1}{e}$ – 2017-01-17
1 Answers
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Let $z-1=t \implies z=t+1$
Therefore, $\frac{e^z}{(z-1)^2}$ = $\frac{e^{(t+1)}}{t^2}=\frac{e}{t^2}.e^t$
=$\frac{e}{t^2}[1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots]$
=$e[\frac{1}{t^2}+\frac{1}{t}+\frac{1}{2!}+\frac{t}{3!}+\frac{t^2}{4!}+\cdots]$
Now, replacing $t$ by $z-1$, we get,
$e[(z-1)^{-2}+(z-1)^{-1}+\frac{1}{2!}+\frac{1}{3!}.(z-1)+\frac{1}{4!}.(z-1)^{2}+\cdots]$
which is the desired expansion.