# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => End of Semester Bonus--sample problem for FE => Topic started by: Victor Ivrii on November 27, 2018, 03:57:31 AM

Title: FE Sample--Problem 5
Post 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 non-real roots?
Title: Re: FE Sample--Problem 5
Post by: Min Gyu Woo 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.

Title: Re: FE Sample--Problem 5
Post by: Victor Ivrii 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 (http://mathworld.wolfram.com/PicardsGreatTheorem.html)
Title: Re: FE Sample--Problem 5
Post by: Min Gyu Woo on November 27, 2018, 11:51:00 AM
Fixed
Title: Re: FE Sample--Problem 5
Post by: Nikita Dua on November 28, 2018, 12:04:42 PM
I don't quite understand why the only root is real?. Can you explain more
Title: Re: FE Sample--Problem 5
Post by: Victor Ivrii 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.