MAT334-2018F > Quiz-7

Q7 TUT 5101

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Victor Ivrii:
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:
$1. f(z)=2z^{4}-2iz^{5}+z^{2}+2iz-1$
$f(x)=2x^{4}-3ix^{3}+x^{2}+2ix-1$
$x: [-\infty, \infty]$
$f(-\infty)\rightarrow \infty$
$f(\infty)\rightarrow \infty$ 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:
$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\ \\ f( iy) \ is\ always\ positive\\ \\ f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\ \\ Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2 \end{array}$

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

Siying Li:
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:
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