Q7 TUT 5301

[Victor Ivrii]

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]

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

[Zechen Wang]

by argument principle,

[Siying Li]

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]

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]

Heng, yes you are correct. I should be more careful to modulus when applying rules on complex numbers. Thank you