MAT334-2018F > Quiz-7

Q7 TUT 0102

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Victor Ivrii:

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.