# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 03:57:02 PM

Title: Q7 TUT 0203
Post by: Victor Ivrii on November 30, 2018, 03:57:02 PM
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given annulus:
$$4z^3- 12z^2 + 2z + 10 \qquad \text{in }\ \bigl\{\frac{1}{2}< |z| < 2\bigr\}.$$
Title: Re: Q7 TUT 0203
Post by: Siying Li on November 30, 2018, 04:13:06 PM
Let$f\left(z\right)\mathrm{=4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\mathrm{+2}z\mathrm{+10}$

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

When$\mathrm{\ }\left|z\right|\mathrm{=}\frac{\mathrm{1}}{\mathrm{2}}$
$\left|\mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\right|\mathrm{=}\left|\mathrm{4}\right|\left|z\right|^{\mathrm{3}}\mathrm{-}\mathrm{12\times }\left|z\right|^{\mathrm{2}}\mathrm{=}\frac{\mathrm{5}}{\mathrm{2}}\mathrm{<}\left|\mathrm{2}\right|\left|\mathrm{z}\right|\mathrm{+10}\mathrm{=}\mathrm{2\times }\frac{\mathrm{1}}{\mathrm{2}}\mathrm{+10}\mathrm{=11}\\ \mathrm{2}z\mathrm{+10=0}\mathrm{\Rightarrow }z\mathrm{=-5}\\ \left|\mathrm{-}\mathrm{5}\right|\mathrm{>}\left|\frac{\mathrm{1}}{\mathrm{2}}\right|\\$
Then $\mathrm{2}z\mathrm{+10}$ has no zero in $\left|z\right|\mathrm{<}\frac{\mathrm{1}}{\mathrm{2}}$

When$\mathrm{\ }\left|z\right|\mathrm{=2}$
$\left|\mathrm{2}z\mathrm{+10}\right|\mathrm{=}\left|\mathrm{2}\right|\left|z\right|+10\mathrm{=14<}\left|\mathrm{4}\right|\left|\mathrm{z}\right|^{\mathrm{3}}\mathrm{-}\mathrm{12}\left|\mathrm{z}\right|^{\mathrm{2}}\mathrm{=}\mathrm{4\times }{\mathrm{2}}^{\mathrm{3}}\mathrm{-}\mathrm{12\times }{\mathrm{2}}^{\mathrm{2}}\mathrm{=16}\\ \mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}\mathrm{=0}\\ \mathrm{4}z^{\mathrm{2}}\left(z\mathrm{-}\mathrm{3}\right)\mathrm{=0}\mathrm{\Rightarrow }\mathrm{4}z^{\mathrm{2}}\mathrm{=0,}z\mathrm{=3}\\ \left|\mathrm{3}\right|\mathrm{>}\left|\mathrm{2}\right|\\$
Then $\mathrm{4}z^{\mathrm{3}}\mathrm{-}\mathrm{12}z^{\mathrm{2}}$ has 2 zeros in $\left|z\right|\mathrm{<2}$

Then number of zeros of f(x) in the region is 2 - 0 = 2