Q7 TUT 0102

[Victor Ivrii]

$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$

Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper half-plane:
$$
2z^4 - 2iz^3 + z^2 + 2iz - 1.
$$

[hz12]

$f\left(z\right)=2z^4-2iz^3+z^2+2iz-1$
When z lies in x axis, z = x + yi = x

So$f\left(x\right)=2x^4-3ix^3+x^2+2ix-1$

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.

[Muyao Chen]

$$f(z) = 2z^{4} - 2iz^{3} + z^{2}+ 2iz -1$$

Consider contour in the upper half-plane 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,$$f(x) = 2x^{4}+x^{2}-1-2ix(x^{2}-1)$$
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 half-plane is 1.

[Yangbo He]

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 x-axis:

if $z = x$
$$f(x) = 2x^4 -2ix^3 + x^2 + 2ix -1= 2x^4 +x^2 -1-2ix(x^2 -1)
$$
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 $-1 < x < \frac{-1-3}{4}, Ref > 0, Imf < 0$, 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.

[Muyao Chen]

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

[Victor Ivrii]

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,

[Victor Ivrii]

Following Yangbo from the place I wrote "WRONG". $2x^4-x^2-1=0$ 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 $4\pi-2\pi=2\pi$ (it must be a multiple of $2\pi$) and there is indeed one root in the upper half-plane. And three in the lower half-plane.

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 half-plane must be purely imaginary and at least one root in the lower half-plane must be purely imaginary. Indeed, consider $g(y):=f(yi)=2y^4-2y^3-y^2-2y-1$ which is real-valued. 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 Wolfram-Alpha to find (approximately) roots of $f(z)$ then Wolfram-Alpha query
Because it is approximate, it thinks that two roots have tiny real parts but we know that they are just imaginary.