Toronto Math Forum

Question: Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $z\cos{\frac{1}{z}}$ at $z = \infty$.

Answer:

Let $w = \frac{1}{z}$, then we are finding Laurent series for $\frac{\cos{w}}{w}$ at $w = 0$.

$\cos{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n}$,

so $z\cos{\frac{1}{z}} = \frac{\cos{w}}{w} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}w^{2n-1} = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n-1}$.

Note that $\frac{\cos{w}}{w} = \frac{1}{w}(1-\frac{w^2}{2!}+\frac{w^4}{4!}-...) = \frac{1}{w}- \frac{w}{2!}+\frac{w^3}{4!}-...$,
so the residue of the function is 1 as the coefficient for the term $\frac{1}{w}$ is 1.

$z\cos{\frac{1}{z}} = \frac{1}{\frac{1}{z}}- \frac{\frac{1}{z}}{2!}+\frac{(\frac{1}{z})^3}{4!}-... = z-\frac{1}{z(2!)}+\frac{1}{z^3(4!)}-...$

$\textbf{Question}$: Find the limit of each function at the given point, or explain why it does not exist: $f(z) = (z-2)log|z-2|$ at $z_0 = 2$.

$\textbf{Answer}$: Since $z_0= 2$, let $z' = z-2$.
$$\lim_{z\to z_0} f(z) = \lim_{z' \to 0} f(z) =z'\log |z'|$$
$$=z' (\ln |z'| +i\cdot 0) = \frac{ln |z'|}{\frac{1}{z'}}$$
By L'Hôpital's Rule, $$= \frac{\frac{1}{z'}}{-\frac{1}{(z')^2}}=\lim_{z'\to 0} (-z') = 0$$.

Therefore, the limit of the function is 0 at $z_0=2$.

According to what has been provided in Textbook: The translation of $\beta$ of $C$ is the circle centered at $z_0+\beta$ with the same radius. If the line L is $\{z:Re(Az+B) = 0\}$, then the translation of L by $\beta$ is the line $\{w: Re(Aw+B-A\beta) = 0\}$

Firstly, we need to find the eigenvalues and eigenvectors.

$det(A-\lambda I) = det\left|\begin{matrix}1-\lambda & 1 & 1 \\ 2 & 1-\lambda & -1 \\ -8 & -5 & -3-\lambda\end{matrix}\right| = (\lambda + 2)(\lambda - 2)(\lambda + 1)$
 
Thus, the eigenvalues are $\lambda_1 = -2, \lambda_2 = -1, \lambda_3 = 2$

When $\lambda_1 = -2, (A - \lambda I) = (A + 2I) = \left(\begin{matrix}3 & 1 & 1 \\ 2 & 3 & -1 \\ -8 & -5 & -1\end{matrix}\right)$

\begin{equation*}
   \begin{pmatrix}
  3 & 1 & 1\\
  2 & 3 & -1\\
  -8 & -5 & -1
  \end{pmatrix}
  \begin{pmatrix}
  x_1\\
  x_2\\
  x_3
  \end{pmatrix}=0
  \end{equation*}
   Let x_1 = t
 \begin{equation*}
\begin{pmatrix}
  x_1\\
  x_2\\
  x_3
  \end{pmatrix}=t
 \begin{pmatrix}
  -4\\
  5\\
  7
  \end{pmatrix}
\end{equation*}

Therefore, the corresponding eigenvector $\vec{v_1} = \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right) $

Similarly, when $\lambda_2 = -1, \lambda_3 = 2$, we can get $\vec{v_2} = \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right)
 and \ \vec{v_3} = \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right)$ respectively.

Therefore, the general solution $X(t) = c_1 e^{-t} \left(\begin{matrix}-3 \\ 4 \\ 2\end{matrix}\right) + c_2 e^{2t} \left(\begin{matrix}0 \\ -1 \\ 1\end{matrix}\right) + c_3 e^{-2t} \left(\begin{matrix}-4 \\ 5 \\ 7\end{matrix}\right)$

Isolate $x_2$ in first equation we get $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$

Differentiate both sides with respect to t we get $x'_2 = \frac{4}{3}x''_1 - \frac{5}{3}x'_1$

Sub into second equation , we get $x''_1 - \frac{5}{2}x'_1 + x_1 = 0$

Characteristic equation is $r^2 - \frac{5}{2} r + 1 = 0$,

hence $r_1 = \frac{1}{2}, r_2 = 2$

General solution for $x_1$ is $x_1 = c_1e^\frac{t}{2}+ c_2 e^{2t}$

Plug into $x_2 = \frac{4}{3}x'_1 - \frac{5}{3}x_1$, we get $x_2 = -c_1 e^\frac{t}{2}+ c_2 e^{2t}$,

So, $x_1 = c_1e^\frac{t}{2}+ c_2e^{2t}$, $x_2 =  -c_1 e^\frac{t}{2} + c_2 e^{2t}$

Plug in $x_1(0) = -2, x_2(0) = 1$,

$c_1 + c_2 = -2, -c_1 + c_2 = 1$

hence $c_1 = -\frac{3}{2}, c_2 = -\frac{1}{2}$

Therefore, $x_1 = -\frac{3}{2} e^\frac{t}{2}-\frac{1}{2} e^{2t}$,
                $x_2 = \frac{3}{2} e^\frac{t}{2} -\frac{1}{2} e^{2t}$

$W(x, xe^x) = \left|\begin{matrix}x & xe^x \\ x' & (xe^x)'\end{matrix}\right|= \left|\begin{matrix}x & xe^x \\ 1 & x^2e^x+e^x\end{matrix}\right| = x(x^2e^x+e^x) - xe^x = x^3e^x$

$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases}
  x = - \frac{p}{\sqrt{p^2+1}} \\
  y = - \frac{1}{p}{x}
 \end{cases}$
No it is not!!! I wrote solution in the parametric form already. Now the question is to exclude $p$ , expressing $y$ via $x$   

Hi Sir, I figured out the problem and corrected the answer below! Thank you!