TUT 0301

[Victor Ivrii]

Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$
f(z)=z^2 + iz + 2 + i.
$$

[Xiaoning Han]

$$x^2+ix+2+i=0 \Rightarrow x=\frac{-\pm\sqrt{-9-4i}}{2}$$
$1)$.
$x\in[0,R]$,
$$\begin{align}f(x)&=x^2+ix+2+i \\&=(x^2+2)(1+\frac{i(x+1)}{x^2+2})\end{align} \\f(0)=2+i\\f(R)=R^2+iR+2+i$$
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.

[Siying Li]

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

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)

[Victor Ivrii]

Completely wrong application of argument principle. And poor typing

[Wanying Zhang]

Sorry.. I think the following one is the correct solution. I had wrong arg computation above.
This one maybe the correct answer.

[Victor Ivrii]

The arguments are correct but two errors, cancelling one another: $\arctan (1/2)\ne \pi /6$ (you could simply write  $\arctan (1/2)$).