1. ## Contour integration

given intergrate from 0 to 2pi

(sin t)^2 dz / jz
---------
5- 4 cos t

i achieve the C_-1 equation till

( z^2 + z^-2 - 2) ( -j )
----------------------- dz
-20z + 8z^2 + 8

The pole are 2 & 0.5. ( Only 0.5 is lying in |z|=1 )

But i can't get the final answer which is ( pi/4) when i sub z =0.5 into

( z^2 + z^-2 - 2) * (2pi)
------------------
-----20+16z

many thanks
ck

2. Originally Posted by Chris0724
given intergrate from 0 to 2pi

(sin t)^2 dz / jz
---------
5- 4 cos t

i achieve the C_-1 equation till

( z^2 + z^-2 - 2) ( -j )
----------------------- dz
-20z + 8z^2 + 8

The pole are 2 & 0.5. ( Only 0.5 is lying in |z|=1 )

But i can't get the final answer which is ( pi/4) when i sub z =0.5 into

( z^2 + z^-2 - 2) * (2pi)
------------------
-----20+16z
many thanks
ck
$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}
{{5 - 4\cos t}}\,dt} = \left\{ \begin{gathered}
{e^{it}} = z, \hfill \\
dt = \frac{{dz}}
{{iz}} \hfill \\
\end{gathered} \right\} = \frac{1}
{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}
{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} .$

$f\left( z \right) = \frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2z - 1} \right)\left( {z - 2} \right)}}.$

So, you have three poles: ${z_{1,2}} = 0,{\text{ }}{z_3} = \frac{1}{2},{\text{ }}{z_4} = 2$. Note that the pole ${z_{1,2}} = 0$ is the double and ${z_4} = 2$ isn't lying in the unit circle $|z|=1$.
Then, according to the residue theorem and Jordan's lemma, you have

$\frac{1}{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} = \frac{1}{{4i}} \cdot 2\pi i\left[ {\mathop {{\text{Res}}}\limits_{z \to 0} \bigr\{ {f\left( z \right)} \bigr\} + \mathop {{\text{Res}}}\limits_{z \to \frac{1}{2}} \bigr\{ {f\left( z \right)} \bigr\}} \right].$

${\color{red}\boxed{\color{black}\begin{gathered}
\mathop {{\text{Res}}}\limits_{z \to a} \bigr\{ {f\left( z \right)} \bigr\} = \frac{1}
{{\left( {n - 1} \right)!}}\mathop {\lim }\limits_{z \to a} \frac{d^{n-1}}
{{d{z^{n - 1}}}}\Bigl[\left(z - a\right)^n f\left( z \right) \Bigr] \hfill \\
a{\text{ is the pole point and }}n{\text{ is the pole multiplicity}} \hfill \\
\end{gathered}}}$

$\mathop {{\text{Res}}}\limits_{z \to 0} \bigr\{ {f\left( z \right)} \bigr\} = \frac{1}
{{\left( {2 - 1} \right)!}}\mathop {\lim }\limits_{z \to 0} \frac{{{d^{2 - 1}}}}
{{d{z^{2 - 1}}}}\left[ {{{\left( {z - 0} \right)}^2}\frac{{{z^4} - 2{z^2} + 1}}
{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}} \right] =$

$= \mathop {\lim }\limits_{z \to 0} \frac{d}
{{dz}}\left[ {\frac{{{z^4} - 2{z^2} + 1}}
{{2{z^2} - 5z + 2}}} \right] = \mathop {\lim }\limits_{z \to 0} \frac{{4{z^5} - 15{z^4} + 10{z^2} - 12z + 8{z^3} + 5}}
{{4{z^4} - 20{z^3} + 33{z^2} - 20z + 4}} = \frac{5}{4}.$

$\mathop {{\text{Res}}}\limits_{z \to \frac{1}{2}} \bigr\{ {f\left( z \right)} \bigr\} = \mathop {\lim }\limits_{z \to \frac{1}{2}} \left( {z - \frac{1}{2}} \right)\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2z - 1} \right) \left( {z - 2} \right)}} = \frac{1}{2}\mathop {\lim }\limits_{z \to \frac{1}{2}} \frac{{{z^4} - 2{z^2} + 1}}{{{z^3} - 2{z^2}}} = - \frac{3}
{4}.$

Finally, you have

$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}{{5 - 4\cos t}}\,dt} = \frac{1}{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} = \frac{1}{{4i}} \cdot 2\pi i\left( {\frac{5}{4} - \frac{3}{4}} \right) = \frac{\pi }{4}.$

3. Originally Posted by DeMath
$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}
{{5 - 4\cos t}}\,dt} = \left\{ \begin{gathered}
{e^{it}} = z, \hfill \\
dt = \frac{{dz}}
{{iz}} \hfill \\
\end{gathered} \right\} = \frac{1}
{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}
{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} .$

$f\left( z \right) = \frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2z - 1} \right)\left( {z - 2} \right)}}.$

So, you have three poles: ${z_{1,2}} = 0,{\text{ }}{z_3} = \frac{1}{2},{\text{ }}{z_4} = 2$. Note that the pole ${z_{1,2}} = 0$ is the double and ${z_4} = 2$ isn't lying in the unit circle $|z|=1$.
Then, according to the residue theorem and Jordan's lemma, you have

$\frac{1}{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} = \frac{1}{{4i}} \cdot 2\pi i\left[ {\mathop {{\text{Res}}}\limits_{z \to 0} \bigr\{ {f\left( z \right)} \bigr\} + \mathop {{\text{Res}}}\limits_{z \to \frac{1}{2}} \bigr\{ {f\left( z \right)} \bigr\}} \right].$

${\color{red}\boxed{\color{black}\begin{gathered}
\mathop {{\text{Res}}}\limits_{z \to a} \bigr\{ {f\left( z \right)} \bigr\} = \frac{1}
{{\left( {n - 1} \right)!}}\mathop {\lim }\limits_{z \to a} \frac{d^{n-1}}
{{d{z^{n - 1}}}}\Bigl[\left(z - a\right)^n f\left( z \right) \Bigr] \hfill \\
a{\text{ is the pole point and }}n{\text{ is the pole multiplicity}} \hfill \\
\end{gathered}}}$

$\mathop {{\text{Res}}}\limits_{z \to 0} \bigr\{ {f\left( z \right)} \bigr\} = \frac{1}
{{\left( {2 - 1} \right)!}}\mathop {\lim }\limits_{z \to 0} \frac{{{d^{2 - 1}}}}
{{d{z^{2 - 1}}}}\left[ {{{\left( {z - 0} \right)}^2}\frac{{{z^4} - 2{z^2} + 1}}
{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}} \right] =$

$= \mathop {\lim }\limits_{z \to 0} \frac{d}
{{dz}}\left[ {\frac{{{z^4} - 2{z^2} + 1}}
{{2{z^2} - 5z + 2}}} \right] = \mathop {\lim }\limits_{z \to 0} \frac{{4{z^5} - 15{z^4} + 10{z^2} - 12z + 8{z^3} + 5}}
{{4{z^4} - 20{z^3} + 33{z^2} - 20z + 4}} = \frac{5}{4}.$

$\mathop {{\text{Res}}}\limits_{z \to \frac{1}{2}} \bigr\{ {f\left( z \right)} \bigr\} = \mathop {\lim }\limits_{z \to \frac{1}{2}} \left( {z - \frac{1}{2}} \right)\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2z - 1} \right) \left( {z - 2} \right)}} = \frac{1}{2}\mathop {\lim }\limits_{z \to \frac{1}{2}} \frac{{{z^4} - 2{z^2} + 1}}{{{z^3} - 2{z^2}}} = - \frac{3}
{4}.$

Finally, you have

$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}{{5 - 4\cos t}}\,dt} = \frac{1}{{4i}}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}{{{z^2}\left( {2{z^2} - 5z + 2} \right)}}\,dz} = \frac{1}{{4i}} \cdot 2\pi i\left( {\frac{5}{4} - \frac{3}{4}} \right) = \frac{\pi }{4}.$

hi DeMath,

Many thanks for the help

i dun understand how you get 3 pole... why is it a must to multiply both numerator & denominator by z^2 ?

4. Originally Posted by Chris0724
hi DeMath,

Many thanks for the help

i dun understand how you get 3 pole... why is it a must to multiply both numerator & denominator by z^2 ?
This is just the usual transformation:

$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}{{5 - 4\cos t}}dt}.$

Numerator: ${\sin ^2}t = {\left( {\frac{{{e^{it}} - {e^{ - it}}}}
{{2i}}} \right)^2} = - \frac{1}
{4}\left( {{e^{2it}} - 2 + \frac{1}
{{{e^{2it}}}}} \right) = - \frac{1}
{{4{e^{2it}}}}\left( {{e^{4it}} - 2{e^{2it}} + 1} \right).$

Denominator: $5 - 4\cos t = 5 - 2\left( {{e^{it}} + {e^{ - it}}} \right) = - \frac{1}{{{e^{it}}}}\left( {2{e^{2it}} - 5{e^{it}} + 2} \right).$

Together: $\frac{{{{\sin }^2}t}}
{{5 - 4\cos t}} = \frac{{ - \frac{1}
{{4{e^{2it}}}}\left( {{e^{4it}} - 2{e^{2it}} + 1} \right)}}
{{ - \frac{1}
{{{e^{it}}}}\left( {2{e^{2it}} - 5{e^{it}} + 2} \right)}} = \frac{1}
{4}\frac{{{e^{4it}} - 2{e^{2it}} + 1}}
{{{e^{it}}\left( {2{e^{2it}} - 5{e^{it}} + 2} \right)}}.$

Then

$\int\limits_0^{2\pi } {\frac{{{{\sin }^2}t}}
{{5 - 4\cos t}}\,dt} = \frac{1}
{4}\int\limits_0^{2\pi } {\frac{{{e^{4it}} - 2{e^{2it}} + 1}}
{{{e^{it}}\left( {2{e^{2it}} - 5{e^{it}} + 2} \right)}}\,dt} = \left\{ \begin{gathered}
{e^{it}} = z, \hfill \\
dt = \frac{{dz}}
{{iz}} \hfill \\
\end{gathered} \right\} =$

$= \frac{1}{4}\oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}
{{z\left( {2{z^2} - 5z + 2} \right)}}\frac{{dz}}{{iz}}} = \frac{1}{{4i}} \oint\limits_{\left| z \right| = 1} {\frac{{{z^4} - 2{z^2} + 1}}{{{z^2} \left( {2{z^2} - 5z + 2} \right)}}\,dz} .$