Author Topic: LEC5101-Quiz7-FIVE-E  (Read 254 times)

RunboZhang

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LEC5101-Quiz7-FIVE-E
« on: December 08, 2020, 07:55:16 PM »
$\textbf{Problem: } \\$
$\text{Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper half-plane.}$ $$2z^4-2iz^3+z^2+2iz-1$$
$\text{(The contour is attached in the picture below)}$

$\textbf{Solution: }\\$
$\text{Denoting the contour } c_1 ,\ c_2 \text{ as indicated in the picture below.}$

$\text{Firstly, discuss contour }c_1 \text{. On }c_1 \text{, we have: }$ $$z=x ,\ x:-R \rightarrow R $$

$\text{And f(z) becomes}$ $$f(z) = 2x^4-2ix^3+x^2+2ix-1=(2x^4+x^2-1)+i(-2x^3+2x)$$

$\text{When } x \in [-R,1] $ $$ Re(f(z))>0 ,\ Im(f(z))<0 ,\ \text{and } \frac{d}{dx}{Re(f(z))}>\frac{d}{dx}{Im(f(z))}$$
$\text{Hence, as Real part is growing much faster than Imaginary part in interval } x \in [-R,1] ,\ arg(f(z)) \rightarrow 0 \text{ in the first quadrant.}$

$\text{Moving forward, as x increases from -1 }$
$$\text{when } x \in (-1, -0.707) ,\ Re>0 ,\ Im<0 \Rightarrow 4^{th} quadrant \\
\text{when } x \in (-0.707,0),\ Re<0 ,\ Im<0  \Rightarrow 3^{rd} quadrant\\
\text{when } x \in (0,0.707),\ Re<0 ,\ Im>0 \Rightarrow 2^{nd} quadrant\\
\text{when } x \in (0.707,1) ,\ Re>0 ,\ Im>0 \Rightarrow 1^{st} quadrant$$

$\text{The change of Real part and Imaginary part makes f(z) travels from forth to third to second to first quadrant.}$

$\text{When } x \in (1, R),\ Re>0 ,\ Im<0 \text{, thus f(z) goes to the } 4^{th} \text{ quadrant.}$

$\text{As } x \rightarrow R ,\ \frac{d}{dx}{Re}>\frac{d}{dx}{Im} \Longrightarrow arg(f(z)) \rightarrow 0 \text{ from the forth quadrant.}$

$\text{Therefore, in contour } c_1 \text{, the total change of argument is } -2\pi \text{ as f(z) rotates clockwise.}$

$\text{Then observe the half-arc contour }c_2 \text{, we have }$ $$z=Re^{it} ,\ t \in (0, \pi)$$

$\text{And f(z) becomes }$
$
\begin{gather}
\begin{aligned}
f(z) &= 2R^{4}e^{4it}-2iR^{3}e^{3it}+R^{2}e^{2it}-1 \\
&=2R^{4}e^{4it}\cdot(1-\frac{ie^{-it}}{R} +\frac{e^{-2it}}{2R^2}-\frac{e^{-4it}}{2R^4}) \\
&\rightarrow 2R^{4}e^{4it} \ \ \text{as } R\rightarrow \infty
\end{aligned}
\end{gather}
$

$\text{Thus as t changes from 0 to }\pi ,\ 2R^{4}e^{4it} \text{ changes from 0 to } 4\pi \text{. And the change of argument in contour } c_2 \text{ is }4\pi$

$\text{Therefore, by the Argument Principle, the number of zeroes equals to }$ $$\frac{1}{2\pi} (-2\pi +4\pi) = 1$$
$\text{only 1 zero.}$
« Last Edit: December 08, 2020, 08:03:03 PM by RunboZhang »