Author Topic: Q7 TUT 0202  (Read 607 times)

Victor Ivrii

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Q7 TUT 0202
« on: November 30, 2018, 03:56:08 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$
f(z)=z^7 + 6z^3 + 7.
$$

Siying Li

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Re: Q7 TUT 0202
« Reply #1 on: November 30, 2018, 04:11:34 PM »
Since $\mathrm{f}\left(\mathrm{z}\right)=z^7+6z^3+7$

Then in the first quadrant,
When z goes from 0 to R on real axis,
$\mathrm{z}\mathrm{=}\mathrm{x}\\\
\mathrm{\ }\mathrm{f}\left(\mathrm{x}\right)=x^7+6x^3+7\\
\ f\left(0\right)=7,{\mathrm{arg} \left(f\left(z\right)\right)\ }=0\\
\mathrm{\ }\mathrm{f}\left(\mathrm{R}\right)=\ +\infty \mathrm{\ }\mathrm{\ }\mathrm{as\ }\mathrm{R\ go}\mathrm{es\ to}\mathrm{+}\mathrm{\infty },{\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}0\\$

When z goes from 0 to iR on imaginary axis,
$\mathrm{z=iy}\\
\mathrm{\ }\mathrm{f}\left(\mathrm{z}\right)={\left(iy\right)}^7+6{\left(iy\right)}^3+7=7-i\left(y^7+6y^3\right)\\
\Re=7,\ {\mathrm{arg} \left(f\left(z\right)\right)\ }=\mathrm{-}\mathrm{arc(}{\mathrm{tan} \left(\frac{y^7+6y^3}{7}\right)\ })\\$
 
When z is in between,
$\mathrm{z=}{\mathrm{R}\mathrm{e}}^{\mathrm{it}},\ 0\le t\le \frac{\pi }{2}\\
\\ f\left(z\right)={\left({\mathrm{R}\mathrm{e}}^{\mathrm{it}}\right)}^7+6{\left({\mathrm{R}\mathrm{e}}^{\mathrm{it}}\right)}^3+7=R^7e^{i7t}+6{R^3e}^{i3t}+7=R^7\left(e^{i7t}+\frac{6e^{i3t}}{R^4}+\frac{7}{R^4}\right)\\
{\mathrm{arg} \left(f\left(z\right)\right)\ }\approx 7t\\
when\ t=0,\ 7t=0\ \\
when\ t=\frac{\pi }{2},\ 7t=2\pi +\frac{3}{2}\pi \\$
 
The net change of argument is  overall $\mathrm{4}\mathrm{\pi}$, so 2 zeros in the first quadrant

sishan

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Re: Q7 TUT 0202
« Reply #2 on: December 01, 2018, 03:11:55 AM »
Let f(z) = u + iv = $z^7 + 6z^3 +7$

Let z = $Re^{i\theta}$, and  $0\leq \theta \leq \frac{\pi}{2}$,  $R\to \infty$

f(z) is analytic at all points except z = $\infty$. Therefore, it is analytic within and upon the complementary of first quadrant.

when z = x,

$f(z) = u + iv = x^7 + 6x^3 + 7$

$arg f = tan^{-1}(\frac{v}{u}) = tan^{-1}(\frac{0}{x^7 + 6x^3 + 7})$ = 0, $\forall$ x $\geq$ 0

Therefore, $arg f = 0$


when z = $Re^{i\theta}$, $0\leq \theta \leq \frac{\pi}{2}$,  $R\to \infty$
 
 f(z) = $R^7e^{7i\theta}(1+\frac{6}{R^4e^{4i\theta}} + \frac{7}{R^7e^{7i\theta}})$
 
 when $R\to \infty$, $f \to R^7e^{7i\theta}$  and arg f = $7\theta$
 
 $argf = 7(\frac\pi2-0) = \frac{7\pi}{2}$


when z = iy,

f(z) = u + iv =$^7 + 6x^3 + 7$

$argf = tan^{-1}(\frac{v}{u})= tan^{-1}(\frac{y^7-6y^3}{7}) = \frac{\pi}{2}$  from $\infty \to 0$



$argf = \frac{7\pi}{2}+\frac{\pi}{2} = 4\pi$

Thus, the angle change is $4\pi$, and the number of zero in the first quadrant is 2.
« Last Edit: December 01, 2018, 03:20:53 AM by sishan »

Victor Ivrii

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Re: Q7 TUT 0202
« Reply #3 on: December 01, 2018, 01:33:26 PM »
everybody is either wrong or missing something

Heng Kan

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Re: Q7 TUT 0202
« Reply #4 on: December 01, 2018, 03:40:52 PM »
Please see the new attached scanned picture.   For yi on Ri to 0, as long as y is positive, f(yi) always lies in the fourth quardarnt.When R tends to be infinity, Re(f(iR)) = 7 and Im(f(iR)) tends to be negtive infinity. When R=0, f(iR) = 7. So f(iy) approximately rotates from negative imaginary axis to positive real axis counter-clockwisely.
« Last Edit: December 01, 2018, 05:44:36 PM by Heng Kan »

Victor Ivrii

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Re: Q7 TUT 0202
« Reply #5 on: December 01, 2018, 04:06:02 PM »
Analysis on straight segments is not complete:

$x$ from $0$ to $R$; indeed, $f(x)$ stays real, but argument of real negative is not $2n\pi $, it is $(2n+1)\pi$. You need to check if the sign of $f(x)$ changes here.

$yi$ from $Ri$ to $0$: there could be an error in the multiple of $2n\pi $. In this case you see that $f(yi)$ is imaginary and does change sign. So one can say: "may be arg changes from $\pi/2$ to $-pi/2$ or may be to $3pi/2$, how do we know?" So you need to look at $f(yi+\varepsilon)$ with $0<\varepsilon\ll 1$ and look how its real part changes signs (or if it changes at all). Use $f(yi+\varepsilon)=f(yi)+ f'(yi) \varepsilon$

Heng Kan

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Re: Q7 TUT 0202
« Reply #6 on: December 01, 2018, 06:01:19 PM »
Please see the new attached scanned picture.   For yi on Ri to 0, as long as y is positive, f(yi) always lies in the fourth quardarnt.When R tends to be infinity, Re(f(iR)) = 7 and Im(f(iR)) tends to be negtive infinity. When R=0, f(iR) = 7. So f(iy) approximately rotates from negative imaginary axis to positive real axis counter-clockwisely.

Victor Ivrii

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Re: Q7 TUT 0202
« Reply #7 on: December 01, 2018, 06:32:03 PM »
Now it is a flawless analysis