Toronto Math Forum

(a) 

Let ${\rm f}\left({\rm z}\right){\rm =}\lambda \frac{a-z}{1-\overline{a}z}$ , $\left|\lambda \right|{\rm =1}$
${\rm where\ }\lambda {\rm =1}{{\rm e}}^{{\rm it}},   {\rm a=}{{\rm re}}^{{\rm i}\theta }$
Then
${\rm f}\left(0\right){\rm =}\lambda \frac{a-0}{1-\overline{a}0}=\lambda a=\frac{1}{2}\ \ \Rightarrow a=\frac{1}{2}e^{i \theta}$
${\rm f}\left({\rm 1}\right){\rm =}{{\rm e}}^{{\rm it}}\frac{\frac{1}{2}e^{i \theta}-1}{1-\frac{1}{2}e^{-i \theta}}\ \Rightarrow \frac{1}{2}-e^{it}=-1+\frac{1}{2}e^{-i \theta}\Rightarrow t= \theta =0$
Thus $\lambda {\rm =1,\ a=}\frac{{\rm 1}}{{\rm 2}}$

Then ${\rm f}\left({\rm z}\right){\rm =}\lambda \frac{a-z}{1-\overline{a}z}=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{1-2z}{2-z}$

(b)
${\rm f}\left({\rm z}\right)=\frac{1-2z}{2-z}=z\ \Rightarrow z^2-4z+1=0\Rightarrow z=2\pm \ \sqrt{3}$
Then the fixed points are $z=2\pm \ \sqrt{3}$

(c)
${\rm f}\left({\rm z}\right){\rm =}\frac{1-2z}{2-z}$
Then
${{\rm f}}^{{\rm '}}\left({\rm z}\right){\rm =}\frac{-2\left(2-z\right)+\left(1-2z\right)}{{\left(2-z\right)}^2}=\frac{-3}{{\left(2-z\right)}^2}$
Then
${{\rm |f}}^{{\rm '}}\left({\rm z}\right)|=\left|\frac{-3}{{\left(2-z\right)}^2}\right|=\frac{3}{{\left(2-z\right)}^2}$
${\arg  \left(f\left(z\right)\right)\ }={\arg  \left(\frac{-3}{(2-z)(2-z)}\right)\ }={\arg \left(-1\right)} - {\arg \frac{{\left(2-z\right)}^2}{3}}={\pi} - {\arg \frac{{\left(2-z\right)}^2}{3}}$

I think the residue at 0 should be 0..
Here's my answer: (in calculation I ignore k for convenience)

${\rm f}\left({\rm z}\right){\rm =}{\tan  \left({\rm z}\right)\ }{\rm +z}{{\cot }^{{\rm 2}} \left(z\right)=\frac{{\sin  \left(z\right)\ }}{{\cos  \left(z\right)\ }}+\frac{z{{\cos }^2 \left(z\right)\ }}{{{\sin }^2 \left(z\right)\ }}\ }=\frac{{{\sin }^3 (z)\ }+z{{\cos }^3 (z)\ }}{{\cos  \left(z\right)\ }{{\sin }^2 \left(z\right)\ }}$
If ${\cos  \left(z\right)\ }=0$, then ${\rm z=}\frac{\pi}{2} + k\pi$

For numerator, ${{\sin }^3 (\frac{\pi}{2})\ }+\frac{\pi}{2}{{\cos }^3 \left(\frac{\pi}{2}\right)\ne 0\ }$, the numerator has zero order of zeros

For denominator, ${\cos  \left(\frac{\pi}{2}\right)\ }{{\sin }^2 \left(\frac{\pi}{2}\right)=0\ ,-{{\sin }^3 \left(\frac{\pi}{2}\right)\ }+\frac{\pi}{2}{{\cos }^2 \left(\frac{\pi}{2}\right)\ }{\sin  \left(\frac{\pi}{2}\right)\ }\ne 0\ }$, the denominator has zeros of order 1

Then ${\rm f}\left({\rm z}\right)$ has simple pole ${\rm z=}\frac{\pi}{2}+k\pi$

If ${{\sin }^{{\rm 2}} \left(z\right)\ }=0$, then ${\rm z=0}$ or ${\rm z=}k\pi $

When ${\rm z=0}$,

For numerator, ${{\sin }^3 (0)\ }+0{{\cos }^3 \left(0\right)=0\ },3{{\sin }^2 \left(0\right)+{{\cos }^3 \left(0\right)\ }+3{{\cos }^2 (0)\ }{\rm sin}?(0)\ }\ \ne 0\ $, the numerator has zero order of 1

For denominator, ${\cos  \left(0\right)\ }{{\sin }^2 \left(0\right)=0\ ,-{{\sin }^3 \left(0\right)\ }+2{{\cos }^2 \left(0\right)\ }{\sin  \left(0\right)\ }=0\ },\ -3{{\sin }^2 \left(0\right){\cos  \left(0\right)\ }-4{\cos  \left(0\right)\ }{{\sin }^{{\rm 2}} \left(0\right)\ }+2cos^3\left(0\right)\ne 0\ }$, the denominator has zeros of order 2

Then ${\rm f}\left({\rm z}\right)$ has simple pole ${\rm z=0}$

When ${\rm z=}k\pi $,

For numerator, ${{\sin }^3 (\pi )\ }+0{{\cos }^3 \left(\pi \right)\ne 0\ }$ the numerator has zero order of 0

For denominator, ${\cos  \left(\pi \right)\ }{{\sin }^2 \left(\pi \right)=0\ ,-{{\sin }^3 \left(\pi \right)\ }+2{{\cos }^2 \left(\pi \right)\ }{\sin  \left(\pi \right)\ }=0\ },\ -3{{\sin }^2 \left(\pi \right){\cos  \left(\pi \right)\ }-4{\cos  \left(\pi \right)\ }{{\sin }^{{\rm 2}} \left(\pi \right)\ }+2cos^3\left(\pi \right)\ne 0\ }$, the denominator has zeros of order 2

Then ${\rm f}\left({\rm z}\right)$ has pole ${\rm z=}k\pi $ with order 2


${\rm Res}\left({\rm f}\left({\rm z}\right),\frac{\pi}{2}+k\pi\right)=\frac{{{\sin }^3 (\frac{\pi }{{\rm 2}})\ }+\frac{\pi}{2}{{\cos }^3 (\frac{\pi}{2})\ }}{-{{\sin }^3 \left(\frac{\pi }{{\rm 2}}\right)\ }+2{{\cos }^2 \left(\frac{\pi }{{\rm 2}}\right)\ }{\sin  \left(\frac{\pi }{{\rm 2}}\right)\ }}=\frac{1}{-1}=-1$
${\rm Res}\left({\rm f}\left({\rm z}\right),k\pi\right)=\frac{3{{\sin }^2 \left(\pi\right)+{{\cos }^3 \left(\pi\right)\ }+3{{\cos }^2 (\pi)\ }{\rm sin}(\pi)\ }}{1!}=\frac{-1}{1}=-1$
Since ${\rm f}\left({\rm z}\right)=\frac{{{\sin }^3 (z)\ }+z{{\cos }^3 (z)\ }}{{\cos  \left(z\right)\ }{{\sin }^2 \left(z\right)\ }}=\frac{{\left(\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n+1\right)!}}\right)}^3+\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n\right)!}}}{\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n}}{\left(2n\right)!}}\ {\left(\int^{\infty }_0{{\left(-1\right)}^n\frac{z^{2n+1}}{\left(2n+1\right)!}}\right)}^2}$ has no term of ${\left({\rm z-0}\right)}^{{\rm -}{\rm 1}}$, then
${\rm Res}\left({\rm f}\left({\rm z}\right),0\right)=0$

Heng, yes you are correct. I should be more careful to modulus when applying rules on complex numbers. Thank you

Correction:
${\mathrm{z}}^{\mathrm{3}}+1\mathrm{=0}$
${\mathrm{z}}^{\mathrm{3}}\mathrm{=}\mathrm{-}\mathrm{1}$
Since $\left|\mathrm{-}\mathrm{1}\right|\mathrm{<}\left|2\right|$, then  ${\mathrm{z}}^{\mathrm{3}}+1$ has 3 zero in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 3 - 1 = 2

Let$\ f\left(z\right)\mathrm{=}z^{\mathrm{3}}\mathrm{-}\mathrm{3}z\mathrm{+1}$

Since $1\mathrm{<}\left|z\right|\mathrm{<2}$

When$\mathrm{\ }\left|z\right|\mathrm{=}1$
$\left|z^{\mathrm{3}}\mathrm{+}\mathrm{1}\right|\mathrm{=}\left|{\mathrm{z}}\right|^{\mathrm{3}}\mathrm{+1}\mathrm{=}\mathrm{2}\mathrm{<}\left|\mathrm{-}\mathrm{3z}\right|\mathrm{=}\left|\mathrm{-}\mathrm{3}\right|\left|z\right|\mathrm{=}\mathrm{3}$

$\mathrm{-}\mathrm{3z}\mathrm{=0}\mathrm{\Rightarrow }z\mathrm{=}0$

$\left|0\right|\mathrm{<}\left|\mathrm{1}\right|$

Then -3z has one zero in $\left|z\right|\mathrm{<}\mathrm{1}$

When$\mathrm{\ }\left|z\right|\mathrm{=2}$
$\left|{\mathrm{z}}^{\mathrm{3}}+1\right|\mathrm{=}\left|\mathrm{z}\right|^{\mathrm{3}}\mathrm{+1}\mathrm{=}\mathrm{9>}\left|\mathrm{-}\mathrm{3z}\right|\mathrm{=}\left|\mathrm{-}\mathrm{3}\right|\left|z\right|\mathrm{=}\mathrm{6}$

${\mathrm{z}}^{\mathrm{3}}+1$ has 3 zero in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 3 - 1 = 2

Since $\mathrm{f}\left(\mathrm{z}\right)=z^7+6z^3+7$

Then in the first quadrant,
When z goes from 0 to R on real axis,
$\mathrm{z}\mathrm{=}\mathrm{x}\\\
\mathrm{\ }\mathrm{f}\left(\mathrm{x}\right)=x^7+6x^3+7\\
\ f\left(0\right)=7,{\mathrm{arg} \left(f\left(z\right)\right)\ }=0\\
\mathrm{\ }\mathrm{f}\left(\mathrm{R}\right)=\ +\infty \mathrm{\ }\mathrm{\ }\mathrm{as\ }\mathrm{R\ go}\mathrm{es\ to}\mathrm{+}\mathrm{\infty },{\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}0\\$

When z goes from 0 to iR on imaginary axis,
$\mathrm{z=iy}\\
\mathrm{\ }\mathrm{f}\left(\mathrm{z}\right)={\left(iy\right)}^7+6{\left(iy\right)}^3+7=7-i\left(y^7+6y^3\right)\\
\Re=7,\ {\mathrm{arg} \left(f\left(z\right)\right)\ }=\mathrm{-}\mathrm{arc(}{\mathrm{tan} \left(\frac{y^7+6y^3}{7}\right)\ })\\$
 
When z is in between,
$\mathrm{z=}{\mathrm{R}\mathrm{e}}^{\mathrm{it}},\ 0\le t\le \frac{\pi }{2}\\
\\ f\left(z\right)={\left({\mathrm{R}\mathrm{e}}^{\mathrm{it}}\right)}^7+6{\left({\mathrm{R}\mathrm{e}}^{\mathrm{it}}\right)}^3+7=R^7e^{i7t}+6{R^3e}^{i3t}+7=R^7\left(e^{i7t}+\frac{6e^{i3t}}{R^4}+\frac{7}{R^4}\right)\\
{\mathrm{arg} \left(f\left(z\right)\right)\ }\approx 7t\\
when\ t=0,\ 7t=0\ \\
when\ t=\frac{\pi }{2},\ 7t=2\pi +\frac{3}{2}\pi \\$
 
The net change of argument is  overall $\mathrm{4}\mathrm{\pi}$, so 2 zeros in the first quadrant