All roots of $e^z=3z^2$ in $B(0,1)$ are real.

Question

I set $f(z):=3z^2$ and $g(z):=-e^z$. Then at $|z|=1$ I get $$ |g(z)| = |-e^z| \le e^{|z|} = e < 3 = 3|z|^2 = |f(z)|, $$ and so by Rouché's theorem $f(z)=3z^2$ and $f(z)+g(z)=3z^2-e^z$ have the same number of roots, counting multiplicites, in the unit disk $B(0,1)$.

My understanding so far is as follows: As $f(z)=3z^2$ has a root of multiplicity $2$ and $z=0$ is a root, $z=0$ must have multiplicity $2$; otherwise, a nonzero imaginary root comes in conjugate pairs which would leave no room for $z=0$ to be a root, and $z \not= 0$ is not a root.

If I am not mistaken so far I think I was able to show that $3z^2$ has only one real root of multiplicity $2$ on the unit disk. But none of this helped me answer the question of showing that the two roots of $3z^2-e^z$ in the unit disk must lie on the real line.

Answer 1

Note that:

• $\exp(-1)<3=3(-1)^2$;
• $\exp(0)=1>3\times0^2$;
• $\exp(1)=e<3\times1^2$.

This proves that the equation $e^z=3z^2$ has at least two real roots on that disk. Together with what you proved, this solves the problem.

Answer 2

I want to change your proof and use Rouché's theorem as you want. Let $\varepsilon<\dfrac{3-e}{2}$ so with $f(z)=-3z^2+\varepsilon$ and $g(z)=e^z-\varepsilon$, in $|z|=1$ $$|g(z)|=|e^z-\varepsilon|\le e^{|z|}+\varepsilon<3-\varepsilon<|3z^2-\varepsilon|=|f(z)|$$ and zeros of $f(z)$ are real in the unit disk $B(0,1)$.