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### Messages - Siying Li

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1
##### Final Exam / Re: FE-P4
« on: December 18, 2018, 01:15:39 PM »
(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
${{\rm |f}}^{{\rm '}}\left({\rm z}\right)|=\left|\frac{-3}{{\left(2-z\right)}^2}\right|=\frac{3}{{\left(2-z\right)}^2}$
${\arg \left(f\left(z\right)\right)\ }={\arg \left(\frac{-3}{(2-z)(2-z)}\right)\ }={\arg \left(-1\right)} - {\arg \frac{{\left(2-z\right)}^2}{3}}={\pi} - {\arg \frac{{\left(2-z\right)}^2}{3}}$

2
##### Final Exam / Re: FE-P2
« on: December 18, 2018, 01:00:53 PM »
Another way to do question (a)

(a)
$\frac{{\rm 1}}{{\rm 2}}\left(z+\frac{1}{z}\right)=\frac{1}{2}\left(e^{logz}+e^{-logz}\right)$
Let ${\rm a=}{\log \left(z\right)\ }$

Then
$\frac{{\rm 1}}{{\rm 2}}\left(z+\frac{1}{z}\right)=\frac{1}{2}\left(e^a+e^{-a}\right)={\rm cosh}(a)={\sin \left(\frac{\pi}{2}+ia\right)\ }={\sin \left(\frac{\pi}{2}\right)\ }{\cosh \left(a\right)\ }+icos\left(\frac{\pi}{2}\right){\rm sinh}?(a)$
Let ${\rm u=}{\sin \left(\frac{\pi}{2}\right)\ }{\cosh \left(a\right)\ }$, ${\rm v=}cos\left(\frac{\pi}{2}\right){\rm sinh}?(a)$

Then ${\sin \left(\frac{\pi}{2}\right)\ }{\rm =}\frac{u}{{\rm cosh}?(a)}\ ,{\cos \left(\frac{\pi}{2}\right)\ }=\frac{v}{{\rm sinh}?(a)}$ , and ${{\sin }^{{\rm 2}} \left(\frac{\pi}{2}\right)\ }+{{\cos }^{{\rm 2}} \left(\frac{\pi}{2}\right)\ }={\left(\frac{u}{{\cosh \left(a\right)\ }}\right)}^2+{\left(\frac{v}{{\sinh \left(a\right)\ }}\right)}^2 =1$

Then ${z:\left|z\right|=r}$ maps to
${w=u+iv:{\left(\frac{u}{{\cosh \left(a\right)\ }}\right)}^2+{\left(\frac{v}{{\sinh \left(a\right)\ }}\right)}^2 =1}$

Where ${\left({\cosh \left({\rm a}\right)\ }\right)}^{{\rm 2}}-{\left({\sinh \left(a\right)\ }\right)}^2=1$

3
##### Final Exam / Re: FE-P3
« on: December 18, 2018, 12:45:02 PM »
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
${\rm Res}\left({\rm f}\left({\rm z}\right),0\right)=0$

4
##### Quiz-7 / Re: TUT 0301
« on: November 30, 2018, 06:14:01 PM »
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{z}}^{\mathrm{2}}+i{\mathrm{z}}+2+i$

When $\mathrm{z\ }$is on Real axis, let $\mathrm{z=x+iy}$, then $\mathrm{y=0}$, $\mathrm{z=x}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{x}}^{\mathrm{2}}-ix+2+i\mathrm{=}{\mathrm{x}}^{\mathrm{2}}+2+i{\left(\mathrm{x+1}\right)}$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{x+1}{x^2+2}\right)\ }$
As R goes to infinity, $\frac{x+1}{x^2+2}$ goes to 0
Then ${\mathrm{arctan} \left(\frac{x+1}{x^2+2}\right)\ }\mathrm{=}0$

When $\mathrm{z\ }$is on Imaginary axis, let $\mathrm{z=x+iy}$, then $\mathrm{x}\mathrm{=0}$, $\mathrm{z=}\mathrm{iy}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{(iy)}}^{\mathrm{2}}+i{\left(iy\right)}+2+i\mathrm{=-}{\mathrm{y}}^{\mathrm{2}}-y+2+i$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{1}{-y^2-y+2}\right)\ }$
As R goes to infinity, $\frac{1}{-y^2-y+2}$ goes to 0
Then ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{1}{-y^2-y+2}\right)\ }\mathrm{=}0$

Let $\mathrm{z=}{\mathrm{Re}}^{\mathrm{it}}\mathrm{,\ \ 0}\mathrm{\le }\mathrm{t}\mathrm{\le }\frac{\mathrm{\pi }}{2}$

Then $\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{({\mathrm{Re}}^{\mathrm{it}})}^{\mathrm{2}}+i{\left({\mathrm{Re}}^{\mathrm{it}}\right)}+2+i\mathrm{=}{\mathrm{R}}^{\mathrm{2}}e^{i2t}+iRe^{it}+2+i$
$\mathrm{arg}\mathrm{}\mathrm{(f}\left(\mathrm{z}\right)\mathrm{)}\mathrm{\cong }\mathrm{2t}$
When $\mathrm{t=0}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=2*0=0}$

When $\mathrm{t=}\frac{\mathrm{\pi }}{2}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=2}\frac{\mathrm{\pi }}{2}\mathrm{=}\mathrm{\pi }$

Then the overall net change in ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }$ is $\left(\mathrm{\pi }\mathrm{-0}\right)\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{=}\mathrm{\pi }$

Then $\frac{1}{2\pi }*\left(\pi \right)\mathrm{=}\frac{1}{2}$
Then the number of zeros of $\mathrm{f}\left(\mathrm{z}\right)$ is 0 (i.e. no zero of f(x) in the first quadrant)

5
##### Quiz-7 / Re: Q7 TUT 5301
« on: November 30, 2018, 05:32:22 PM »
Heng, yes you are correct. I should be more careful to modulus when applying rules on complex numbers. Thank you

6
##### Quiz-7 / Re: Q7 TUT 5101
« on: November 30, 2018, 05:28:32 PM »
Solve with argument principle

$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{z}}^{\mathrm{2}}-2+{\mathrm{i}}$

When $\mathrm{z\ }$is on Real axis, let $\mathrm{z=x+iy}$, then $\mathrm{y=0}$, $\mathrm{z=x}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{z}}^{\mathrm{2}}-2+{\mathrm{i}}\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{x}}^{\mathrm{2}}-2+{\mathrm{i}}$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{3x^2+1}{4x^4+x-2}\right)\ }$
As R goes to infinity, $\frac{3x^2+1}{4x^4+x-2}\mathrm{=}0$
Then ${\mathrm{arctan} \left(\frac{3x^2+1}{4x^4+x-2}\right)\ }\mathrm{=}0$

Let $\mathrm{z=}{\mathrm{Re}}^{\mathrm{it}}\mathrm{,\ \ 0}\mathrm{\le }\mathrm{t}\mathrm{\le }\mathrm{\pi }$
Then $\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{({\mathrm{Re}}^{\mathrm{it}})}^{\mathrm{4}}+3i{\left({\mathrm{Re}}^{\mathrm{it}}\right)}^2-2+i\mathrm{=4}{\mathrm{R}}^{\mathrm{4}}e^{i4t}+3iR^2e^{i2t}-2+i$
$\mathrm{arg}\mathrm{}\mathrm{(f}\left(\mathrm{z}\right)\mathrm{)}\mathrm{\cong }\mathrm{4t}$
When $\mathrm{t=0}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4*0=0}$

When $\mathrm{t=}\mathrm{\pi }$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4}\ \mathrm{\pi }\mathrm{=4}\mathrm{\pi }$

Then the overall net change in ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }$ is $\left(\mathrm{4}\mathrm{\pi }\mathrm{-0}\right)\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{=4}\mathrm{\pi }$

Then the number of zeros in $\mathrm{f}\left(\mathrm{z}\right)$ is $\frac{1}{2\pi }*\left(4\pi \right)=2$

7
##### Quiz-7 / Re: Q7 TUT 5301
« on: November 30, 2018, 04:41:03 PM »
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

8
##### Quiz-7 / Re: Q7 TUT 5201
« on: November 30, 2018, 04:36:29 PM »
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}-3{\mathrm{z}}^{\mathrm{2}}+3$

When $\mathrm{z\ }$is on Real axis, let $\mathrm{z=x+iy}$, then $\mathrm{y=0}$, $\mathrm{z=x}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}-3x^2+3\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}-3x^2+3$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{0}{4x^4-3x^2+3}\right)\ }=0$

When $\mathrm{z\ }$is on Imaginary axis, let $\mathrm{z=x+iy}$, then $\mathrm{x}\mathrm{=0}$, $\mathrm{z=}\mathrm{iy}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{\mathrm{(iy)}}^{\mathrm{4}}-3{\left(iy\right)}^2+3\mathrm{=4}{\mathrm{y}}^{\mathrm{4}}+3y^2+3$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{0}{\mathrm{4}{\mathrm{y}}^{\mathrm{4}}+3y^2+3}\right)\ }=0$

Let $\mathrm{z=}{\mathrm{Re}}^{\mathrm{it}}\mathrm{,\ \ 0}\mathrm{\le }\mathrm{t}\mathrm{\le }\frac{\mathrm{\pi }}{2}$

Then $\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{({\mathrm{Re}}^{\mathrm{it}})}^{\mathrm{4}}-3{\left({\mathrm{Re}}^{\mathrm{it}}\right)}^2+3\mathrm{=4}{\mathrm{R}}^{\mathrm{4}}e^{i4t}-3R^2e^{i2t}+3$
$\mathrm{arg}\mathrm{}\mathrm{(f}\left(\mathrm{z}\right)\mathrm{)}\mathrm{\cong }\mathrm{4t}$
When $\mathrm{t=0}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4*0=0}$

When $\mathrm{t=}\frac{\mathrm{\pi }}{2}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4}\frac{\mathrm{\pi }}{2}\mathrm{=2}\mathrm{\pi }$

Then the overall net change in ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }$ is $\left(\mathrm{2}\mathrm{\pi }\mathrm{-0}\right)\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{=2}\mathrm{\pi }$

Then the number of zeros in $\mathrm{f}\left(\mathrm{z}\right)$ is $\frac{1}{2\pi }*\left(2\pi \right)=1$

*Corrected typo $\mathrm{\pi }$ as $\frac{\mathrm{\pi }}{2}$, thank you

9
##### Quiz-7 / Re: Q7 TUT 5301
« on: November 30, 2018, 04:32:59 PM »
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

10
##### Quiz-7 / Re: Q7 TUT 0203
« on: November 30, 2018, 04:13:06 PM »
Let$f\left(z\right)\mathrm{=4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\mathrm{+2}z\mathrm{+10}$

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

When$\mathrm{\ }\left|z\right|\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}$
$\left|\mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\right|\mathrm{=}\left|\mathrm{4}\right|\left|z\right|^{\mathrm{3}}\mathrm{-}\mathrm{12\times }\left|z\right|^{\mathrm{2}}\mathrm{=}\frac{\mathrm{5}}{\mathrm{2}}\mathrm{<}\left|\mathrm{2}\right|\left|\mathrm{z}\right|\mathrm{+10}\mathrm{=}\mathrm{2\times }\frac{\mathrm{1}}{\mathrm{2}}\mathrm{+10}\mathrm{=11}\\ \mathrm{2}z\mathrm{+10=0}\mathrm{\Rightarrow }z\mathrm{=-5}\\ \left|\mathrm{-}\mathrm{5}\right|\mathrm{>}\left|\frac{\mathrm{1}}{\mathrm{2}}\right|\\$
Then $\mathrm{2}z\mathrm{+10}$ has no zero in $\left|z\right|\mathrm{<}\frac{\mathrm{1}}{\mathrm{2}}$

When$\mathrm{\ }\left|z\right|\mathrm{=2}$
$\left|\mathrm{2}z\mathrm{+10}\right|\mathrm{=}\left|\mathrm{2}\right|\left|z\right|+10\mathrm{=14<}\left|\mathrm{4}\right|\left|\mathrm{z}\right|^{\mathrm{3}}\mathrm{-}\mathrm{12}\left|\mathrm{z}\right|^{\mathrm{2}}\mathrm{=}\mathrm{4\times }{\mathrm{2}}^{\mathrm{3}}\mathrm{-}\mathrm{12\times }{\mathrm{2}}^{\mathrm{2}}\mathrm{=16}\\ \mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\mathrm{=0}\\ \mathrm{4}z^{\mathrm{2}}\left(z\mathrm{-}\mathrm{3}\right)\mathrm{=0}\mathrm{\Rightarrow }\mathrm{4}z^{\mathrm{2}}\mathrm{=0,}z\mathrm{=3}\\ \left|\mathrm{3}\right|\mathrm{>}\left|\mathrm{2}\right|\\$
Then $\mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}$ has 2 zeros in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 2 - 0 = 2

11
##### Quiz-7 / Re: Q7 TUT 0202
« on: November 30, 2018, 04:11:34 PM »
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

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