TT2B Problem 2

[Victor Ivrii]

(a) Find the decomposition into power series at ${z=0}$ of $$f(z)=(1-z)^{-1}.$$ What is the radius of convergence?

(b) Plugging in $-z^2$ instead of $z$ and integrating, obtain a decomposition at $z=0$ of  $\arctan (z)$.

[Yifei Wang]

I think the power is -1/2 inside of the -1

Correct V.I.  It was actually Test2B

[Wanying Zhang]

Here is the solution to problem 2.

You need to know decomposition of $(1-z)^{-1}$. The rest is simply wrong. V.I.

[Huanglei Ln]

$$
   \begin{aligned}   
    a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=0}z^n \\
   \frac{1}{R}&=\lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1
    \end{aligned}
$$
$$
   \begin{aligned}   
    b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\
   \Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\
    \Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\
    \Rightarrow \arctan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\
    \Rightarrow \arctan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c
    \end{aligned}
$$

[Huanglei Ln]

 \begin{aligned}
    a)f(z)&=\frac{1}{1-z}=\sum^{\infty}_{n=o}z^n \\
 \frac{1}{R}&=lim_{n\rightarrow {\infty} }|\frac{1}{1}|=1\Rightarrow R=1
  \end{aligned}
 \end{displaymath}
\begin{displaymath}
 \begin{aligned}
    b)f(-z^2)&=\frac{1}{1+z^2}=\sum^{\infty}_{n=o}{-z^2}^n=\Sigma^{\infty}(-1)^nz^{2n} \\
 \Rightarrow \int f(-z^2)dz&=\sum^{\infty}_{n=o}(-1)^n\int^{2n}dz\\
    \Rightarrow \int \frac{1}{1+z^2}dz&=\sum^{\infty}_(n=0)(-1)^n\int z^{2n}dz\\
    \Rightarrow artan(z)+c&=\sum^{\infty}_{n=0}(-1)^n \frac{z^{2n+1}}{2n+1}\\
    \Rightarrow artan(z)&=\sum^{\infty}_{n=0}\frac{(-1)^n}{2n+1}z^{2n+1} +c
  \end{aligned}

[Victor Ivrii]

Huanglei

I fixed your LaTeX. Don't learn it from crappy sources!

Also as $z=0$ you'll see that $c=0$. In actual test missing this will lead to the mark reduction