Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz7 => Topic started by: Victor Ivrii 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:
$$
f(z)=z^7 + 6z^3 + 7.
$$

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,
$\mathrm{z=iy}\\
\mathrm{\ }\mathrm{f}\left(\mathrm{z}\right)={\left(iy\right)}^7+6{\left(iy\right)}^3+7=7i\left(y^7+6y^3\right)\\
\Re=7,\ {\mathrm{arg} \left(f\left(z\right)\right)\ }=\mathrm{}\mathrm{arc(}{\mathrm{tan} \left(\frac{y^7+6y^3}{7}\right)\ })\\$
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

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 = $7\theta$
$argf = 7(\frac\pi20) = \frac{7\pi}{2}$
when z = iy,
f(z) = u + iv =$^7 + 6x^3 + 7$
$argf = tan^{1}(\frac{v}{u})= tan^{1}(\frac{y^76y^3}{7}) = \frac{\pi}{2}$ 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.

everybody is either wrong or missing something

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 counterclockwisely.

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 $f(yi+\varepsilon)=f(yi)+ f'(yi) \varepsilon$

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 counterclockwisely.

Now it is a flawless analysis