Toronto Math Forum
MAT334-2018F => MAT334--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 03:59:32 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:
$$
f(z)=z^4 - 3z^2 + 3.
$$
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$f\left(z\right)=z^4-3z^2+3$
When z lines in x axis, z = x+ yi = x
So $f\left(x\right)=x^4-3x^2+3$
Because the domain of f(x) is $\left[0,\left.R\right]\right.$, so arg(f(z)) = 0
Let $z=Re^{it}$, when $0\le t\le \frac{\pi }{2}$,
So f($Re^{it}$) = $R^4e^{4it}-3R^2e^{2it}+3$, $\mathrm{0}\mathrm{\le }\mathrm{4}\mathrm{t}\le \2pi $,
Hence arg(f(z)) = 2$\pi$
When z = yi $0\le y\le R$ arg(f(z)) = 0
So the net change of the angle is 0 + 2$\pi$ + 0 = 2$\pi$, and $\frac{1}{2\pi }\bullet 2pi\ =1$
There are 1 zero of the function.
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here is my solution
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$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}-3{\mathrm{z}}^{\mathrm{2}}+3$
When $\mathrm{z\ }$is on Real axis, let $\mathrm{z=x+iy}$, then $\mathrm{y=0}$, $\mathrm{z=x}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}-3x^2+3\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}-3x^2+3$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{0}{4x^4-3x^2+3}\right)\ }=0$
When $\mathrm{z\ }$is on Imaginary axis, let $\mathrm{z=x+iy}$, then $\mathrm{x}\mathrm{=0}$, $\mathrm{z=}\mathrm{iy}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{\mathrm{(iy)}}^{\mathrm{4}}-3{\left(iy\right)}^2+3\mathrm{=4}{\mathrm{y}}^{\mathrm{4}}+3y^2+3$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{0}{\mathrm{4}{\mathrm{y}}^{\mathrm{4}}+3y^2+3}\right)\ }=0$
Let $\mathrm{z=}{\mathrm{Re}}^{\mathrm{it}}\mathrm{,\ \ 0}\mathrm{\le }\mathrm{t}\mathrm{\le }\frac{\mathrm{\pi }}{2}$
Then $\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{({\mathrm{Re}}^{\mathrm{it}})}^{\mathrm{4}}-3{\left({\mathrm{Re}}^{\mathrm{it}}\right)}^2+3\mathrm{=4}{\mathrm{R}}^{\mathrm{4}}e^{i4t}-3R^2e^{i2t}+3$
$\mathrm{arg}\mathrm{}\mathrm{(f}\left(\mathrm{z}\right)\mathrm{)}\mathrm{\cong }\mathrm{4t}$
When $\mathrm{t=0}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4*0=0}$
When $\mathrm{t=}\frac{\mathrm{\pi }}{2}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4}\frac{\mathrm{\pi }}{2}\mathrm{=2}\mathrm{\pi }$
Then the overall net change in ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }$ is $\left(\mathrm{2}\mathrm{\pi }\mathrm{-0}\right)\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{=2}\mathrm{\pi }$
Then the number of zeros in $\mathrm{f}\left(\mathrm{z}\right)$ is $\frac{1}{2\pi }*\left(2\pi \right)=1$
*Corrected typo $\mathrm{\pi }$ as $\frac{\mathrm{\pi }}{2}$, thank you
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the domain of t should be 0<=t<=2/π, not π
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$f(z) = z^{4} -3z^{2} + 3$ substitute $w = z^{2}$
$$ z^{4} -3z^{2} + 3 = w^{2} - 3w + 3 = 0$$
$ w = \frac{3 \pm i \sqrt 3}{2}$, so $ z=\sqrt{\frac{3 \pm i \sqrt 3}{2}}$
For z in $[0, R]$:
$$f(x)= x^{4}-3x^{2}+ 3$$
$f(0) = 3$, then $arg(f(z)) = 0$
For $z = Re^{it}$ And $0 \leq t \leq \frac{\pi}{2}$:
$$f(Re^{it}) = R^{4}e^{4it} - eRe^{2it} + 3 = R^{4}(e^{4it} - \frac{3e^{2it}}{R^{3}} + \frac{3}{R^{4}}) = R^{4}e^{4it}(1 - \frac{3}{R^{3} e^{2it}} +\frac{3}{R^{4}}) $$
As $R \rightarrow \infty$
$f(z)= R^{4}e^{4it}$ along 2, so that t goes from 0 to $\frac{\pi}{2}$, then $arg(f(z))$ goes from $4 *0 = 0$ to $4* \frac{\pi}{2}$ = 2$\pi$
For $z = iy$, And $0 \leq y \leq R$:
$$f(iy) = y^{4} - 3y + 3$$
$$f(0) = 3$$
Then $arg(f(z)) = 0$
Then $$ \triangle arg(f(z)) = 0 +2 \pi + 0 = 2 \pi$$
$$\frac{1}{2 \pi}[\triangle arg(f(z))] = N_{0} - N_{p} = N_{0} = \frac{1}{2 \pi}2 \pi = 1$$
Then only one solution of the function in the first quadrant.
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All solutions are wrong even if some answers are correct: When saying $\arg(f(+\infty)) =0$ and $f(0)=0$ you forgot that $\arg$ is defined modulo $2\pi n$. You must trace change of $\arg$ carefully
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Since in [0, positive infinity) f(x) is always positive and always stays on the real axis, the change in arg on [0, R] should be zero.
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Since in [0, positive infinity) f(x) is always positive and always stays on the real axis, the change in arg on [0, R] should be zero.
It is a good start. What about $[Ri,0]$? And about arc (well someone wrote about arc, but I want everything in one place)
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here is the solution. since there are no changes along both positive axis, change of arg is zero on those lines. The only change of arg is on the arc = 2pi.
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So, it is how this problem should be done: calculating the change of $\arg f(z)$ along circular arc and also along straight segment(s). In this rather simple case $f(z)$ is real and does not change sign along straight lines; so the corresponding increment of $\arg f(z)$ is $0$.