Author Topic: Q7 TUT 5301  (Read 597 times)

Victor Ivrii

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Q7 TUT 5301
« on: November 30, 2018, 04:00:08 PM »
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given annulus:
$$
z^3- 3z+1 \qquad \text{in   } \bigl\{1< |z| < 2\bigr\}.
$$

Siying Li

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Re: Q7 TUT 5301
« Reply #1 on: November 30, 2018, 04:32:59 PM »
Let$\ f\left(z\right)\mathrm{=}z^{\mathrm{3}}\mathrm{-}\mathrm{3}z\mathrm{+1}$

Since $1\mathrm{<}\left|z\right|\mathrm{<2}$

When$\mathrm{\ }\left|z\right|\mathrm{=}1$
$\left|z^{\mathrm{3}}\mathrm{+}\mathrm{1}\right|\mathrm{=}\left|{\mathrm{z}}\right|^{\mathrm{3}}\mathrm{+1}\mathrm{=}\mathrm{2}\mathrm{<}\left|\mathrm{-}\mathrm{3z}\right|\mathrm{=}\left|\mathrm{-}\mathrm{3}\right|\left|z\right|\mathrm{=}\mathrm{3}$

$\mathrm{-}\mathrm{3z}\mathrm{=0}\mathrm{\Rightarrow }z\mathrm{=}0$

$\left|0\right|\mathrm{<}\left|\mathrm{1}\right|$

Then -3z has one zero in $\left|z\right|\mathrm{<}\mathrm{1}$

When$\mathrm{\ }\left|z\right|\mathrm{=2}$
$\left|{\mathrm{z}}^{\mathrm{3}}+1\right|\mathrm{=}\left|\mathrm{z}\right|^{\mathrm{3}}\mathrm{+1}\mathrm{=}\mathrm{9>}\left|\mathrm{-}\mathrm{3z}\right|\mathrm{=}\left|\mathrm{-}\mathrm{3}\right|\left|z\right|\mathrm{=}\mathrm{6}$

${\mathrm{z}}^{\mathrm{3}}+1$ has 3 zero in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 3 - 1 = 2
« Last Edit: November 30, 2018, 08:59:21 PM by Siying Li »

Zechen Wang

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Re: Q7 TUT 5301
« Reply #2 on: November 30, 2018, 04:40:14 PM »
by argument principle,
« Last Edit: November 30, 2018, 05:12:52 PM by Zechen Wang »

Siying Li

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Re: Q7 TUT 5301
« Reply #3 on: November 30, 2018, 04:41:03 PM »
Correction:
${\mathrm{z}}^{\mathrm{3}}+1\mathrm{=0}$
${\mathrm{z}}^{\mathrm{3}}\mathrm{=}\mathrm{-}\mathrm{1}$
Since $\left|\mathrm{-}\mathrm{1}\right|\mathrm{<}\left|2\right|$, then  ${\mathrm{z}}^{\mathrm{3}}+1$ has 3 zero in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 3 - 1 = 2

Heng Kan

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Re: Q7 TUT 5301
« Reply #4 on: November 30, 2018, 05:00:10 PM »
Si Ying, I think there is somthing wrong in your solution, although your answer is correct.  You cannot say |z^3+1|=|1^3+1| when
|z| = 1 and |z^3+1| = |2^3+1| when |z|=2 because z^3 is not the same as 1^3 or 2^3. They just have the same modulus.
See my answer on the scanned picture.

Siying Li

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Re: Q7 TUT 5301
« Reply #5 on: November 30, 2018, 05:32:22 PM »
Heng, yes you are correct. I should be more careful to modulus when applying rules on complex numbers. Thank you