# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Final Exam => Topic started by: Victor Ivrii on December 18, 2018, 06:11:23 AM

Title: FE-P1
Post by: Victor Ivrii on December 18, 2018, 06:11:23 AM
(a) Decompose into Taylor series at $0$ function $$f(z)=\frac{1}{z^2+2z+2}.$$ Find the radius of convergence $r$. Determine if the series is converging at $|z|=r$ (consider all points $z$ satisfying $|z|=r$).

(b) Decompose into Laurent's series at $\infty$ the same function. Also find the radius $R$ (so it converges as $|z|> R$).
Determine if the series is converging at $|z|=R$ (consider all points $R$ satisfying $|z|=R$).

Hint: Represent $f(z)$ as the sum of functions of the form $\frac{a}{b+z}$.

Title: Re: FE-P1
Post by: Qi Zeng on December 18, 2018, 07:16:16 AM
\begin{align*}
f(z) &= \frac{1}{(z-(-1+i))(z-(-1-i))}\\
&= -\frac{1}{2}i\frac{1}{z-(-1+i)} + \frac{1}{2}i\frac{1}{z-(-1-i)}\\
\end{align*}

(a)
\begin{align*}
f(z) &= -\frac{i}{2(-1+i)}\times\frac{1}{\frac{z}{(-1+i)}-1} + \frac{i}{2(-1-i)}\times\frac{1}{\frac{z}{(-1-i)}-1}\\
&=\frac{i}{2(-1+i)}\times \sum\limits_{n=0}^{\infty}(\frac{z}{-1+i})^n - \frac{i}{2(-1-i)}\times\sum\limits_{n=0}^{\infty}(\frac{z}{-1-i})^n
\end{align*}

The function converges when $|\frac{z}{-1+i}| < 1$ and $|\frac{z}{-1-i}| < 1$, therefore $\frac{|z|}{\sqrt{2}} < 1$, and $|z| < \sqrt{2}$.

The radius of convergence r = $\sqrt{2}$, and it diverges when $|z|$ = $\sqrt{2}$.

(b)
\begin{align*}
f(z) &= -\frac{i}{2z}\times\frac{1}{1 - \frac{-1+i}{z}} + \frac{i}{2z}\times\frac{1}{1 - \frac{-1-i}{z}}\\
&=-\frac{i}{2z}\times \sum\limits_{n=0}^{\infty}(\frac{-1+i}{z})^n + \frac{i}{2z}\times\sum\limits_{n=0}^{\infty}(\frac{-1-i}{z})^n
\end{align*}

The function converges when $|\frac{-1+i}{z}| < 1$ and $|\frac{-1-i}{z}| < 1$, therefore $\frac{\sqrt{2}}{|z|} < 1$, and $|z| > \sqrt{2}$.

The radius R = $\sqrt{2}$, and it diverges when $|z|$ = $\sqrt{2}$.
Title: Re: FE-P1
Post by: hz12 on December 18, 2018, 10:47:02 AM
f(z)=$\frac{1}{z^2+2z+2}$
Let $z^2+2z+2=0,$ we can have $\ {(z+1)}^2=i^2,{\ z}_1=i-1,\ z_2=-i-1$
let $f\left(z\right)=\ \frac{1}{z^2+2z+2}=\ \frac{A}{z-\left(i+1\right)}+\frac{B}{z-\left(-i-1\right)}$
= $\frac{A\left(z-\left(-i+1\right)\right)+B(z-\left(i+1\right))}{(z-\left(i+1\right))(z-\left(-i+1\right))}$
=$\frac{\left(A+B\right)z-A-B+\left(A-B\right)i}{(z-\left(i+1\right))(z-\left(-i+1\right))}$
After calculation, we can have $A=-i/2,\ B=i/2$
So f(Z) = $\frac{-i}{2}\bullet \frac{1}{z-\left(i+1\right)}+\frac{i}{2}\bullet \frac{1}{Z-(-i+1)}$
a,      f(z) = $-\frac{i}{2}\bullet \frac{1}{i+1}\bullet \frac{1}{\frac{z}{i+1}-1}+\frac{i}{2}\bullet \frac{1}{-i+1}\bullet \frac{1}{\frac{z}{-i+1}-1}$
= $\frac{i}{2(i+1)}\bullet \sum^{\infty }_{n=0}{{(\frac{z}{i+1})}^n-\frac{i}{2(1-i)}\sum^{\infty }_{n=0}{{(\frac{z}{-i+1})}^n}}$

For convergence, we need $\left|\frac{z}{i+1}\right|<1,\ \left|\frac{z}{-i+1}\right|<1,\ \ so\ we\ can\ have\ z<\sqrt{2}.$

b,       f(z) = $-\frac{i}{2}\cdot \frac{1}{z}\cdot \frac{1}{1-\frac{i+1}{z}}$ + $\frac{i}{2}\cdot \frac{1}{Z}\cdot \frac{1}{1-\frac{-i+1}{z}}$
= $\frac{i}{2z}\cdot \sum^{\infty }_{n=0}{{(\frac{i+1}{z})}^n-\frac{i}{2z}\sum^{\infty }_{n=0}{{(\frac{1-i}{z})}^n}}$
For convergence, we need  $\left|\frac{i+1}{z}\right|<1,\ \left|\frac{1-i}{z}\right|<1,\ so\ we\ can\ have\ z>\sqrt{2}$
When z =$\sqrt{2}$ , for $\sum^{\infty }_{n=0}{a_n},\ {\mathop{\mathrm{lim}}_{n\to \infty } a_n\neq 0\ }$, so we can conclude it is geometric divergent.
Title: Re: FE-P1
Post by: Muyao Chen on December 18, 2018, 01:46:33 PM
$$f(z) = \frac{A}{z-(-1+i)} + \frac{B}{z-(-1-i)}$$

Solve A, B
$$A = -\frac{i}{2}, B = \frac{i}{2}$$

Then
$$f(z) = -\frac{i}{2}\frac{1}{z-(-1+i)} +\frac{i}{2} \frac{1}{z-(-1-i)}$$

When $\mid z \mid = r$

$$f(z) = -\frac{i}{2} \frac{1}{-1+i} \frac{1}{\frac{z}{-1+i}-1} + \frac{i}{2}\frac{1}{-1-i}\frac{1}{\frac{z}{-1-i}-1}$$
$$= \frac{i}{2} \frac{1}{-1+i} \frac{1}{1 - \frac{z}{-1+i}} - \frac{i}{2}\frac{1}{-1-i}\frac{1}{1 - \frac{z}{-1-i}}$$
$$= \frac{i}{2} \frac{1}{-1+i} \sum_{n=0}^{\infty} (\frac{z}{-1+i})^{n} - \frac{i}{2}\frac{1}{-1-i}\sum_{n=0}^{\infty}(\frac{z}{-1-i})^{n}$$

So that converge at
$$\mid \frac{z}{-1+i} \mid < 1$$
$$\mid z \mid < \sqrt{2}$$

So that not converge at $\mid z \mid = \sqrt{2}$

When $\mid z \mid = R$
$$f(z) = -\frac{i}{2} \frac{1}{z} \frac{1}{ 1 - \frac{-1+i}{z}} + \frac{i}{2}\frac{1}{z}\frac{1}{1 - \frac{-1-i}{z}}$$
$$= \frac{i}{2} \frac{1}{z} \sum_{n=0}^{\infty} (\frac{-1+i}{z})^{n} + \frac{i}{2}\frac{1}{z}\sum_{n=0}^{\infty}(\frac{-1-i}{z})^{n}$$

So that converge at
$$\mid \frac{-1+i}{z} \mid < 1$$
$$\mid z \mid > \sqrt{2}$$

So that not converge at $\mid z \mid = \sqrt{2}$
Title: FE-P1 official
Post by: Victor Ivrii on December 20, 2018, 04:41:00 AM
Observe that
$$f(z)=\frac{1}{(z+1+i)(z+1-i)}= \frac{1}{2i}\Bigl(\frac{1}{z-z_2}-\frac{1}{z-z_1}\Bigr)$$
has two singular points $z_{1,2} =-1\pm i=\sqrt{2}e^{\pm 5\pi i/4}$.

(a)   So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z_2-z}-\frac{1}{z_1-z}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^n z_2^{-n-1} - \sum_{n=0}^\infty z^n z_1^{-n-1}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^n \Bigr(z_2^{-n-1} - z_1^{-n-1}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^n 2^{-(n+1)/2}\Bigl(e^{5(n+1)\pi i/4} - e^{-5(n+1)\pi i/4}\Bigr)=\\
& \sum_{n=0}^\infty 2^{-(n+1)/2}\sin (5(n+1))\pi /4)z^n
\end{align*}
As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.

(b) So $R=\sqrt{2}$ and
\begin{align*}
f(z)=
&\frac{1}{2i}\Bigl(\frac{1}{z-z_1}-\frac{1}{z-z_2}\Bigr)=
\frac{1}{2i}\Bigl(  \sum_{n=0}^\infty z^{-n-1} z_1^{n} - \sum_{n=0}^\infty z^{-n-1} z_2^{n}\Bigr)=\\
&\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} \Bigr(z_1^{n} - z_2^{n}\Big)=
\frac{1}{2i} \sum_{n=0}^\infty z^{-n-1} 2^{n/2}\Bigr(e^{5n\pi i/4} - e^{-5n\pi i/4}\Big)=\\
&\sum_{n=0}^\infty 2^{n/2}\sin (5n\pi /4)z^{-n-1} = \sum_{-\infty}^{-2} 2^{-(n+1)/2}\sin (5z(-n-1)\pi /4)z^{n}
\end{align*}
where we plugged $n=m-1$ with $m=1,2,\ldots$, observed that the term with $m=1$ is $0$ and finally replaced $m$ by $n$.

As $|z|=\sqrt{2}$ terms do not tend to $0$ and the series diverges.