Q6 TUT 0102

[Victor Ivrii]

Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$
f(z)=\frac{z}{\sin^2(z)};\qquad z_0=0\quad \text{(four terms of the Laurent series)} .
$$

[wentao huang]

my solution would be showed in the scan photos.

[Muyao Chen]

$\frac{z}{(sinz)^2}$
sinz = 0, Z = 0
numerator f(z)= z, f(0) = 0,  f'(z) $\neq$ 0, so order = 1
g(z)= $(sinz)^2$, g(0)= 0, g'(z) = 2sinzcosz = sin2z,
g''(z) = 2cos2z $\neq$ 0, so order = 2.
so it's simple pole
Then
$\frac{z}{(sinz)^2}$ = $a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...$
$\frac{z}{(Z - \frac{Z^{3}}{3!} +\frac{Z^{5}}{5!} -...)^2}$ = $a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...$
$Z = (Z - \frac{Z^{3}}{3!} +\frac{Z^{5}}{5!} -...)^2(a_{-1}Z^{-1} + a_{0} +a_{1}Z + a_{2}Z^{2}+ ...)$
Then
$\frac{z}{(sinz)^2}$ =  $\frac{1}{z}$ + $\frac{z}{3}$ + $\frac{z^{3}}{15}$ + $\frac{2Z^{5}}{189}$ +...
$Res(f;0)= 1$

[Qi Cui]

$$let\ h(z) = z, g(z) = {sin(z)}^2$$
$As \ z _0 = 0:$
$$h(0) = 0, h'(0)= 1 \ne1\quad\therefore order = 1$$
$$g(z) = {sin(z)}^2 = 0$$
$$g'(z) = 2sin(z)cos(z), g'(0) = 0$$
$$g''(z) = −2(sin^2(z)-cos^2(z)), g''(0) = 2\ne0, \quad\therefore order = 2$$
$$\quad\therefore 2-1= 1   \quad\therefore \ order\ of \ pole = 1$$
$$\quad\therefore{{z}\over sin^{2}(z)} = a_{-1}z^{-1} + a_{0} + a_{1}z^{1} +...$$
$$z= (z-{z^{3}\over 3!} + {z^{5}\over 5!} -... )^{2}(a_{-1}z^{-1} + a_{0} + a_{1}z^{1} + ...  ) $$
$After\ equating\ coefficients\ of\ equal\ powers\ of\ z, we\ can\ get:$
$$a_{-1}=1,a_{0}=0, a_{1}={{1}\over{3}}, a_{2}= 0, a_{3}={{1}\over{15}},a_{4}=0,a_{5}= {{2}\over{189}}$$

$$\quad\therefore {{1}\over {z}} + {{z}\over3} + {{1z^3}\over15} +{{2z^5}\over{189}} +..., Res(f;0)=1$$

[Victor Ivrii]

Wentao, how many time I requested not to  post crappy photos. Scan! Reading this is not good for eyesight.

Qi, Muyao, you need to be more explicit with calculation of coefficients: need to write equations.