Q7 TUT 0201

[Victor Ivrii]

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

[Yuechen Huang]

Answer is attached.

[hz12]

Since $ze^z-\frac{1}{4}=0\ {{\mathop{\Leftrightarrow}\limits_{}}}\ 4z-e^{-z}=0$
$\ f\left(z\right)=4z,\ s\left(z\right)=e^{-z},\ for\ 0<\left|z\right|<2.$
When $\left|z\right|=2,\ \left|s(z)\right|=\left|e^{-z}\right|=e^{-Re(z)}\le e^2\cong 7.387\dots <8$

And                   $\left|4z\right|=4\left|z\right|=4\bullet 2=8$

So $\left|s(z)\right|<\left|f(z)\right|,\ \ for\ \left|z\right|=2.$

Hence $g\left(z\right)=f\left(z\right)-s\left(z\right)=4z-e^{-z}$ has the same number of zeros

As $f\left(z\right)=4z\ \ in\ \left|z\right|<2$, this is 1 zero.

And when z = 0,$g\left(0\right)=4\bullet 0-e^0=-1\neq 0$

Hence z = 0 is not a zero of g(z).

We can conclude that $ze^z-\frac{1}{4}=0\ has\ 1\ zero\ in\ 0<\left|z\right|<2$.

[Muyao Chen]

$$p(z) = ze^{z} - \frac{1}{4}$$
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 $0 < \mid z \mid < 2$.

[Yangbo He]

This is the answer i got:

[Victor Ivrii]

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