MAT334-2018F > Quiz-7

Q7 TUT 5201

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Victor Ivrii:
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:
$f\left(z\right)=z^4-3z^2+3$
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:
here is my solution

Siying Li:
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}-3{\mathrm{z}}^{\mathrm{2}}+3$

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

Zechen Wang:
the domain of t should be 0<=t<=2/π, not π

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