### Author Topic: Q7 TUT 5201  (Read 1424 times)

#### Victor Ivrii

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##### Q7 TUT 5201
« on: November 30, 2018, 03:59:32 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the first quadrant:
$$f(z)=z^4 - 3z^2 + 3.$$

#### hz12

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##### Re: Q7 TUT 5201
« Reply #1 on: November 30, 2018, 04:13:22 PM »
$f\left(z\right)=z^4-3z^2+3$
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.

#### Zechen Wang

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##### Re: Q7 TUT 5201
« Reply #2 on: November 30, 2018, 04:14:17 PM »
here is my solution

#### Siying Li

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##### Re: Q7 TUT 5201
« Reply #3 on: November 30, 2018, 04:36:29 PM »
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}-3{\mathrm{z}}^{\mathrm{2}}+3$

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
« Last Edit: November 30, 2018, 05:16:37 PM by Siying Li »

#### Zechen Wang

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##### Re: Q7 TUT 5201
« Reply #4 on: November 30, 2018, 04:42:44 PM »
the domain of t should be 0<=t<=2/π, not π

#### Muyao Chen

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##### Re: Q7 TUT 5201
« Reply #5 on: November 30, 2018, 11:32:41 PM »
$f(z) = z^{4} -3z^{2} + 3$ 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.

« Last Edit: December 01, 2018, 12:36:32 AM by Muyao Chen »

#### Victor Ivrii

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##### Re: Q7 TUT 5201
« Reply #6 on: December 01, 2018, 01:21:59 PM »
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

#### Zechen Wang

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##### Re: Q7 TUT 5201
« Reply #7 on: December 01, 2018, 02:22:06 PM »
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.

#### Victor Ivrii

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##### Re: Q7 TUT 5201
« Reply #8 on: December 01, 2018, 03:24:02 PM »
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)

#### Zechen Wang

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##### Re: Q7 TUT 5201
« Reply #9 on: December 01, 2018, 04:58:00 PM »
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.

#### Victor Ivrii

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 $0$.