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MAT334-2018F => MAT334--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 03:57:53 PM

Title: TUT 0301
Post by: Victor Ivrii on November 30, 2018, 03:57:53 PM
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$f(z)=z^2 + iz + 2 + i.$$
Title: Re: TUT 0301
Post by: Xiaoning Han on November 30, 2018, 03:58:15 PM
$$x^2+ix+2+i=0 \Rightarrow x=\frac{-\pm\sqrt{-9-4i}}{2}$$
$1)$.
$x\in[0,R]$,
\begin{align}f(x)&=x^2+ix+2+i \\&=(x^2+2)(1+\frac{i(x+1)}{x^2+2})\end{align} \\f(0)=2+i\\f(R)=R^2+iR+2+i
Thus, $\arg(f(x))$ starts from $\arctan(\frac{1}{2})$ and ends on $\arctan(\frac{R+1}{R^2+2})$.
$\\$
$2)$.
$x=Re^{it},0\leq t\leq \frac{\pi}{2}$,
\begin{align}f(Re^{it})&=(Re^{it})^2+i(Re^{it})+2+i \\&=R^2(e^{it}-\frac{ie^{it}}{R}+\frac{2+i}{R^2})\end{align}, 0\leq t \leq \pi.
$\\$
$3)$.
$x=iy, 0 \leq y \leq R$,
$$f(iy)=-y^2-y+2+i \\ y=0 \Rightarrow 2+i \\ y=R \Rightarrow -R^2-R+2+i$$
Thus, it is a horizontal line.
$\\$
Therefore, the net changes of the argument is $0$, there are no zeros in the first quadrant.
Title: Re: TUT 0301
Post by: Siying Li 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)
Title: Re: TUT 0301
Post by: Victor Ivrii on December 01, 2018, 01:14:15 PM
Completely wrong application of argument principle. And poor typing
Title: Re: TUT 0301
Post by: Wanying Zhang on December 01, 2018, 07:31:54 PM
Sorry.. I think the following one is the correct solution. I had wrong arg computation above.
This one maybe the correct answer.
Title: Re: TUT 0301
Post by: Victor Ivrii on December 01, 2018, 08:08:01 PM
The arguments are correct but two errors, cancelling one another: $\arctan (1/2)\ne \pi /6$ (you could simply write  $\arctan (1/2)$).