Deon Moran

2022-10-24

Where the method used should be using complex analysis.

$${\int}_{c}\frac{d\theta}{(p+\mathrm{cos}\theta {)}^{2}}=\frac{2\pi p}{({p}^{2}-1)\sqrt{{p}^{2}-1}};c:\left|z\right|=1$$

$${\int}_{c}\frac{d\theta}{(p+\mathrm{cos}\theta {)}^{2}}=\frac{2\pi p}{({p}^{2}-1)\sqrt{{p}^{2}-1}};c:\left|z\right|=1$$

Kash Osborn

Beginner2022-10-25Added 18 answers

i do my self like this on $|z|=1,z={e}^{i\theta},d\theta =\frac{dz}{iz}$

using substitution $\mathrm{cos}\theta =\frac{z+{z}^{-1}}{2}$

$\frac{1}{i}\int \frac{\frac{dz}{z}}{(p+\frac{z+{z}^{-1}}{2})(p+\frac{z+{z}^{-1}}{2})}or\frac{1}{i}\int \frac{1}{(2pz+{z}^{2}+1)(p+\frac{z+{z}^{-1}}{2})}$

and then multiple by $\frac{z}{z}$

$\frac{1}{i}\int \frac{z}{(2pz+{z}^{2}+1)(2pz+{z}^{2}+1)}$

the roots of $(2pz+{z}^{2}+1)$, ${z}_{1}=-p+\sqrt{{p}^{2}-1};{z}_{2}=-p-\sqrt{{p}^{2}-1}$

let ${z}_{1}=a$ and ${z}_{2}=b$

so $a-b=2\sqrt{{p}^{2}-1}$ and $a+b=-2p$

and then i use residue theorm, but the result on the right $\frac{\pi p}{2({p}^{2}-1)\sqrt{{p}^{2}-1}}$

using substitution $\mathrm{cos}\theta =\frac{z+{z}^{-1}}{2}$

$\frac{1}{i}\int \frac{\frac{dz}{z}}{(p+\frac{z+{z}^{-1}}{2})(p+\frac{z+{z}^{-1}}{2})}or\frac{1}{i}\int \frac{1}{(2pz+{z}^{2}+1)(p+\frac{z+{z}^{-1}}{2})}$

and then multiple by $\frac{z}{z}$

$\frac{1}{i}\int \frac{z}{(2pz+{z}^{2}+1)(2pz+{z}^{2}+1)}$

the roots of $(2pz+{z}^{2}+1)$, ${z}_{1}=-p+\sqrt{{p}^{2}-1};{z}_{2}=-p-\sqrt{{p}^{2}-1}$

let ${z}_{1}=a$ and ${z}_{2}=b$

so $a-b=2\sqrt{{p}^{2}-1}$ and $a+b=-2p$

and then i use residue theorm, but the result on the right $\frac{\pi p}{2({p}^{2}-1)\sqrt{{p}^{2}-1}}$

Christopher Saunders

Beginner2022-10-26Added 6 answers

On $|z|=1$, we have $z={e}^{i\theta}$, so $\frac{dz}{d\theta}=i{e}^{i\theta}$, and hence $d\theta =\frac{dz}{iz}$

Also on |z|=1, we have $\mathrm{cos}\theta =\frac{1}{2}(z+\frac{1}{z})$

Putting this together, you get

$$\frac{1}{i}{\int}_{|z|=1}\frac{1}{z{(p+\frac{1}{2}(z+\frac{1}{z}))}^{2}}\text{}dz.$$

Expand the denominator,

$$\begin{array}{rl}\frac{1}{i}{\int}_{|z|=1}\frac{1}{z{(p+\frac{1}{2}(z+\frac{1}{z}))}^{2}}\text{}dz& =\frac{1}{i}{\int}_{|z|=1}\frac{1}{z({p}^{2}+p(z+1/z)+\frac{1}{4}(z+1/z{)}^{2})}\\ & =\frac{1}{i}{\int}_{|z|=1}\frac{1}{({p}^{2}z+p{z}^{2}+p+\frac{z}{4}({z}^{2}+2+1/{z}^{2}))}\\ & =\cdots \end{array}$$

Also on |z|=1, we have $\mathrm{cos}\theta =\frac{1}{2}(z+\frac{1}{z})$

Putting this together, you get

$$\frac{1}{i}{\int}_{|z|=1}\frac{1}{z{(p+\frac{1}{2}(z+\frac{1}{z}))}^{2}}\text{}dz.$$

Expand the denominator,

$$\begin{array}{rl}\frac{1}{i}{\int}_{|z|=1}\frac{1}{z{(p+\frac{1}{2}(z+\frac{1}{z}))}^{2}}\text{}dz& =\frac{1}{i}{\int}_{|z|=1}\frac{1}{z({p}^{2}+p(z+1/z)+\frac{1}{4}(z+1/z{)}^{2})}\\ & =\frac{1}{i}{\int}_{|z|=1}\frac{1}{({p}^{2}z+p{z}^{2}+p+\frac{z}{4}({z}^{2}+2+1/{z}^{2}))}\\ & =\cdots \end{array}$$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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