Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz7 => Topic started by: Victor Ivrii on November 30, 2018, 03:53:54 PM

Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given
annulus disk:
$$
ze^z\frac{1}{4} \qquad \text{in }\ \bigl\{0< z < 2\bigr\}.
$$

Answer is attached.

Since $ze^z\frac{1}{4}=0\ {{\mathop{\Leftrightarrow}\limits_{}}}\ 4ze^{z}=0$
$\ f\left(z\right)=4z,\ s\left(z\right)=e^{z},\ for\ 0<\leftz\right<2.$
When $\leftz\right=2,\ \lefts(z)\right=\lefte^{z}\right=e^{Re(z)}\le e^2\cong 7.387\dots <8$
And $\left4z\right=4\leftz\right=4\bullet 2=8$
So $\lefts(z)\right<\leftf(z)\right,\ \ for\ \leftz\right=2.$
Hence $g\left(z\right)=f\left(z\right)s\left(z\right)=4ze^{z}$ has the same number of zeros
As $f\left(z\right)=4z\ \ in\ \leftz\right<2$, this is 1 zero.
And when z = 0,$g\left(0\right)=4\bullet 0e^0=1\neq 0$
Hence z = 0 is not a zero of g(z).
We can conclude that $ze^z\frac{1}{4}=0\ has\ 1\ zero\ in\ 0<\leftz\right<2$.

$$p(z) = ze^{z}  \frac{1}{4}$$
Since $$f(0) \neq 0$$
It would be same as finding the number of zeros in
$$\mid z \mid < 2$$
On
$$\mid z \mid = 2$$
$$\mid ze^{z}\mid = 2e^{Re(z)} > 2e^{2} = 0.276 > \frac{1}{4}$$
So p(z) and $ze^{z}$ have the same number of zero in $\mid z \mid < 2$.
So that number of zeros of f(z)is one in $0 < \mid z \mid < 2$.

This is the answer i got:

Hanyu is correct , Yuechen writing $z=2\implies e^{z}=e$" is wrong