Toronto Math Forum
MAT3342018F => MAT334Tests => End of Semester Bonussample problem for FE => Topic started by: Victor Ivrii 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 nonreal roots?

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.

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 (http://mathworld.wolfram.com/PicardsGreatTheorem.html)

Fixed

I don't quite understand why the only root is real?. Can you explain more

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.