Question

$F(z)=\dfrac{1}{1-z}$, when expanded as a power series around $z=2$, would result in $F(z)=\sum_{k=0}^{\infty} a_k (z-2)^k$ with ROC $|z-2|<1$. The coefficients $a_k,\;k\ge 0$, are given by the expression ________.

• A. $(-1)^k$
• B. $(-1)^{k+1}$
• C. $\left(\dfrac{1}{2}\right)^k$
• D. $\left(-\dfrac{1}{2}\right)^{k+1}$

Hint:

Around a nonzero center $z_0$, rewrite the function in powers of $w=z-z_0$ and apply the geometric series $\frac{1}{1-u}=\sum u^k$ when $|u|<1$.

Answer 1

The Correct Option is B

Step 1: Shift to $(z-2)$. Let $w=z-2 \Rightarrow z=2+w$. Then \[ F(z)=\frac{1}{1-(2+w)}=\frac{1}{-1-w}=-\frac{1}{1+w}. \] Step 2: Use the geometric series for $|w|<1$. \[ \frac{1}{1+w}=\sum_{k=0}^{\infty}(-1)^k w^k \Rightarrow -\frac{1}{1+w}=\sum_{k=0}^{\infty}(-1)^{k+1} w^k. \] Since $w=z-2$, we have $a_k=(-1)^{k+1}$ and ROC $|z-2|=|w|<1$.