Question

Taylor's series expansion of $f\left(z\right)=\frac{z-1}{z+1}$ about the point $z=0$ is

• A. $1+2(z+z^2+z^3\dots \dots )$
• B. $-1-2(z-z^2+z^3\dots \dots )$
• C. $-1+2(z-z^2+z^3\dots \dots )$
• D. None

Answer 1

The correct option is C $-1+2(z-z^2+z^3\dots \dots )$

$f\left(z\right)=\frac{z-1}{z+1}=1-\frac{2}{z+1}$
$\Rightarrow f\left(0\right)=-1,f\left(1\right)=0$
$\Rightarrow f^{{}^{\prime }}\left(z\right)=\frac{2}{(z+1{)}^2}{f}^{{}^{\prime }}\left(0\right)=2$
$f^{\prime \prime }\left(z\right)=-\frac{4}{(z+1{)}^3}{f}^{\prime \prime }\left(0\right)=-4$
$f^{{}^{\prime \prime \prime }}\left(z\right)=\frac{12}{(z+1{)}^4}{f}^{{}^{\prime \prime \prime }}\left(0\right)=12$ and so on
Taylor series: $f\left(z\right)=f\left(z_0\right)+(z-z_0)f^{{}^{\prime }}\left(z_0\right)+\frac{(z-z_0{)}^2}{2!}{f}^{\prime \prime }\left(z_0\right)+\frac{(z-z_0{)}^3}{3!}{f}^{{\prime \prime }^{\prime }}\left(z_0\right)+\dots$
$f\left(z\right)=-1+z\left(2\right)+\frac{z^2}{2}(-4)+\frac{z^3}{6}\left(12\right)+\dots \dots$
$=-1+2z-2z^2+2z^3\dots \dots$
$f\left(z\right)=-1+2(z-z^2+z^3+\dots \dots )$