Author Topic: FE Sample--Problem 5  (Read 1572 times)

Victor Ivrii

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FE Sample--Problem 5
« on: November 27, 2018, 03:57:31 AM »
Show that the equation
$$
e^{z}=e^2z
$$
has a real root in the unit disk $\{z\colon |z|<1\}$.

Are there non-real roots?
« Last Edit: November 27, 2018, 07:12:05 AM by Victor Ivrii »

Min Gyu Woo

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Re: FE Sample--Problem 5
« Reply #1 on: November 27, 2018, 10:54:10 AM »
Let $f(z) = e^2z$ and $g(z) = e^z$.

We have

$|f(z)| = e^2 > e = |g(z)| \text{ when} |z|<1$

So, $f(z) = 0 $ has the same number of zeros as $f(z)=g(z)$. (Q16 in textbook)


$$f(z) = e^2z=0$$
$$z = 0$$

Thus, $f(z)$ has one zero $\implies$ $f(z)=g(z)$ has one zero.

Let's look at the case where z is real. I.e. $z = x$ where $x\in\mathbb{R}$

Then,

Call $h(z) = f(z) - g(z) = e^x - e^2 x$

Note that:

$$h(0) = 1 $$

$$h(1) = e - e^2 <0 $$

By Mean Value Theorem, there exists $x$ where $0 < x < 1$ such that $h(x) = 0$.

I.e. there is a REAL ROOT x where $0<x<1$.

We know that there is only one real root in $\{z:|z|<1\}$ because $|h(\overline{z_0})| = |h(z_0)| =0$ is only true when $z_0$ is real.

Thus, within $|z| <1$ there is only one root, and that root is real.

« Last Edit: November 28, 2018, 01:27:30 PM by Min Gyu Woo »

Victor Ivrii

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Re: FE Sample--Problem 5
« Reply #2 on: November 27, 2018, 11:17:04 AM »
That the only root in $\{z\colon |z|<1\}$ is real follows from the fact that if $z$ is a root, then $\bar{z}$ it also a root. However the last statement do you mean $\{z\colon |z|<1\}$ ? Otherwise it is wrong due to Picard great theorem
« Last Edit: November 27, 2018, 11:25:16 AM by Victor Ivrii »

Min Gyu Woo

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Re: FE Sample--Problem 5
« Reply #3 on: November 27, 2018, 11:51:00 AM »
Fixed

Nikita Dua

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Re: FE Sample--Problem 5
« Reply #4 on: November 28, 2018, 12:04:42 PM »
I don't quite understand why the only root is real?. Can you explain more

Victor Ivrii

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Re: FE Sample--Problem 5
« Reply #5 on: November 30, 2018, 04:02:42 AM »
We know that there is just one root. Min Gyu proved that there exists a real root.

Alternatively we can see that if $z$ is the root, so is $\bar{z}$, which would not be equal $z$ if $z$ is not real. But then we would have at least two roots.