Author Topic: Q7 TUT 5101  (Read 511 times)

Victor Ivrii

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Q7 TUT 5101
« on: November 30, 2018, 03:58:52 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper half-plane:
$$
z^4 + 3iz^2 + z - 2 + i.
$$

Yuechen Huang

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Re: Q7 TUT 5101
« Reply #1 on: November 30, 2018, 04:12:50 PM »
$1. f(z)=2z^{4}-2iz^{5}+z^{2}+2iz-1$
$f(x)=2x^{4}-3ix^{3}+x^{2}+2ix-1$
$x: [-\infty, \infty]$
 $f(-\infty)\rightarrow \infty$
 $f(\infty)\rightarrow \infty$ arg$(f(z))=0$

2. $Re^{it}$   $0\leq t \leq \pi$
$f(t)=2R^{4}e^{i4t}- 2iR^{3}e^{i3t}+R^{2}e^{i2t}+2iRe^{it}-1$
 $0\leq 4t \leq 4\pi$   arg$(f(z))=4\pi$

The net change of argument is $4\pi$ , so that there are four solutions

Jeffery Mcbride

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Re: Q7 TUT 5101
« Reply #2 on: November 30, 2018, 04:24:14 PM »
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\
\\
f( iy) \ is\ always\ positive\\
\\
f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\
\\
Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2
\end{array}$

This is positive at x = R, then negative when it hits x = 1 to x = -1 then it becomes positive again as x approaches -R
.
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$ This is always positive.

So, we are always in the upper half plane. Change in Arg = 0.

$\displaystyle f\left( Re^{it}\right) \ =\ e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term domindates. So the change in argument is 4$\displaystyle \pi $.

$\displaystyle \frac{0+4\pi }{2\pi } \ =\ 2$ zeros total

 
« Last Edit: December 01, 2018, 09:37:25 PM by Jeffery Mcbride »

Siying Li

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Re: Q7 TUT 5101
« Reply #3 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 $\frac{1}{2\pi }*\left(4\pi \right)=2$

Victor Ivrii

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Re: Q7 TUT 5101
« Reply #4 on: December 01, 2018, 01:09:35 PM »
Siying, Yuechen,

Why are you bringing here other problems? It is called offtop and some can it spam .

Jeffery,
If you plug in $z=yi$ you will get not real valued , so "$f(iy)$ 𝑖is always positive" is both incorrect and irrelevant. But something (tell us what) is along $(-\infty, \infty)$ and this would save the day


Jeffery Mcbride

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Re: Q7 TUT 5101
« Reply #5 on: December 01, 2018, 04:54:19 PM »
You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \  >\ 0$ 

When $\displaystyle z\ \rightarrow \ \pm \infty $, Re(f(z)) $\displaystyle \rightarrow \ \infty $ and Im(f(z)) $\displaystyle \rightarrow \ \infty \ $and the endpoints of $\displaystyle f([ -R,\ R]) \ $lie in the first quadrant for R really big. So the change in argument is going to be 0.

This can be coupled with what I said about the curve.

$\displaystyle f\left( Re^{it}\right)$ =  $\displaystyle e^{4it} \ $from [0, $\displaystyle \pi $] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi $.

So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi $.

« Last Edit: December 01, 2018, 08:49:48 PM by Jeffery Mcbride »

Victor Ivrii

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Re: Q7 TUT 5101
« Reply #6 on: December 01, 2018, 05:51:45 PM »
Yes, $f(x)$ belongs to upper half-plane for $x\in (-\infty,\infty)$. But why?

Jeffery Mcbride

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Re: Q7 TUT 5101
« Reply #7 on: December 01, 2018, 08:49:16 PM »
Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \  >\ 0$ 

(Added this to my previous answer)