Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz7 => Topic started by: Victor Ivrii on November 30, 2018, 03:52:38 PM

$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper halfplane:
$$
2z^4  2iz^3 + z^2 + 2iz  1.
$$

$f\left(z\right)=2z^42iz^3+z^2+2iz1$
When z lies in x axis, z = x + yi = x
So$f\left(x\right)=2x^43ix^3+x^2+2ix1$
Because the domain of f(x) is $\left[\infty ,\infty \right]$, $f\left(\infty \right)\to \infty ,\ f\left(\infty \right)\to \infty ,\ $so arg(f(z)) = 0
Let $z=Re^{it}$, , when 0 $\mathrm{<}$= t $\mathrm{<}$= $\pi$
$f\left(t\right)=2R^4e^{i4t}2iR^3e^{i3t}+R^2e^{i2t}+2iRe^{it}1$
And $0\le 4t\le 4\pi $
So arg(f(z)) = 4$\pi$
So the net change of the angle is $0+4\pi =4\pi $ , and $\frac{1}{2\pi }\bullet 4\pi =2$
There are 2 zeroes of the function.

$$f(z) = 2z^{4}  2iz^{3} + z^{2}+ 2iz 1$$
Consider contour in the upper halfplane with radius R.
let $$z = Re^{i \theta}$$
with$$\theta \in [0, \pi]$$
Then$$f(Re^{i \theta}) = R^{4}( (2e^{4i \theta}) + O( \frac{1}{R} ))$$
Then$$argf(Re^{i \theta}) = 4 \pi$$
On the real axis,$$f(x) = 2x^{4}+x^{2}12ix(x^{2}1)$$
Then zeros for the real part is:$$x = \frac{1 \pm 3}{4}$$
for the imaginary part is:$$x = \pm 1, 0$$
When $$x < 1 $$it's in first quadratic.
When $$1 < x < \frac{1 \pm 3}{4} $$it's moved to fourth quadrant.
So that$$\triangle argf(x) = 2 \pi$$
Then$$\frac{1}{2 \pi} (\triangle argf(z)) = 1$$
So that calculate the number of zeroes of the following function in the upper halfplane is 1.

Let f(z) = $2z^4 2iz^3 +z^2 + 2iz 1$
Let z = $Re^{i\theta}$, and $0\leq \theta \le \pi$ where $R\to \infty$
$f(Re^{i\theta}) = 2(Re^{i\theta})^4 2i(Re^{i\theta})^3 +(Re^{i\theta})^2 + 2i(Re^{i\theta}) 1$
$f(Re^{i\theta}) = R^4e^{4i\theta}(2 \frac{2i}{Re^{i\theta}} + \frac{1}{R^2e^{2i\theta}} +\frac{2}{R^3e^{3i\theta}}  \frac{1}{R^4e^{4i\theta}})$
$\triangle arg f(Re^{i\theta}) = 4\pi$ This was a standard part V.I.
On the xaxis:
if $z = x$
$$f(x) = 2x^4 2ix^3 + x^2 + 2ix 1= 2x^4 +x^2 12ix(x^2 1)
$$
The real part is $\Re(f(x))=2x^4 +x^2 1$, and the imaginary part is $\Im(f(x)) =2x(x^2 1)$.
As for the imaginary part = 0, $x = \pm 1, 0$.
As for the real part = 0, $x = \frac{1\pm 3}{4}$. WRONG
If $x < 1, Re f > 0, Imf > 0$, in the first quadrant.
If $1 < x < \frac{13}{4}, Ref > 0, Imf < 0$, in the fourth quadrant.
Therefore, $\triangle argf(x)\arrowvert_{\gamma x} = 2\pi$
$N  P = \frac{1}{2\pi}\triangle argf(z)\arrowvert_{\gamma} = \frac{\triangle argf(x)\arrowvert_{\gamma x} + \triangle argf(Re^{i\theta})\arrowvert_{\gamma R}}{2\pi} = 1$
The number of zero is 1.

Because as R $\rightarrow \infty$, others goes to 0.

Muyao's idea is correct but you need to be more specific where is $f(x)$ for different values of $x$. For that you need also find where $\Im(f(x))$ changes sign,

Following Yangbo from the place I wrote "WRONG". $2x^4x^21=0$ is biquadratic equation and we hget $x^=\frac{1}{2},1$ and real roots are $\pm\frac{1}{\sqrt{2}}$. So all roots of $\Re f(x)$, $\Im f(x)$ differ and are simple, $f(x)$ does not vanish and $\Re(f(x))$ or $\Im (f(x))$ change sign when $x$ passes the corresponding value. Thus the following scheme holds (see picture) and increment of $\arg (f(x))$ is $\approx 2\pi$ (the error tends to $0$ as $R\to \infty$) and thus the increment of $\arg (f(z))$ along this pass is $4\pi2\pi=2\pi$ (it must be a multiple of $2\pi$) and there is indeed one root in the upper halfplane. And three in the lower halfplane.
Addition: observe that $\overline{f(z)}=f(\bar{z})$ and therefore $z$ is a root implies that $\bar{z}$ (its symmetric about imaginary axis) is a root as well. Therefore one root in the upper halfplane must be purely imaginary and at least one root in the lower halfplane must be purely imaginary. Indeed, consider $g(y):=f(yi)=2y^42y^3y^22y1$ which is realvalued. As $y\to \pm \infty$ it tends to $+\infty$ and $f(0)=1$ so it must have real roots as $y>0$ and as $y<0$. As $y>0$ it has just one root since $f(z)$ has just one root there (we are talking about roots of real polynomial $g(y)$).
As $y<0$ there is one real root of $g(y)$. Indeed: we see that $g'(y)= 8y^3 6y^2 2y 2<0$ for $y<0$.
Let us ask WolframAlpha to find (approximately) roots of $f(z)$ then WolframAlpha query (https://www.wolframalpha.com/input/?i=Solve++++2x%5E4%E2%88%922ix%5E3%2Bx%5E2%2B2ix%E2%88%921%3D0)
Because it is approximate, it thinks that two roots have tiny real parts but we know that they are just imaginary.