Toronto Math Forum
MAT334-2018F => MAT334--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 03:56:08 PM
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Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$
f(z)=z^7 + 6z^3 + 7.
$$
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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
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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.
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everybody is either wrong or missing something
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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.
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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 $f(yi+\varepsilon)=f(yi)+ f'(yi) \varepsilon$
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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.
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Now it is a flawless analysis