how to show whether there are non-real roots in the unit disc of this equation?

Question

For the equation $e^z=e^2z$, $|z|\leq 1$.

I have shown there are no roots on the imaginary axis and boundary of the unit disc by simple computing.

And let $z=x+iy$ ($x^2+y^2\leq 1$). $e^z=e^2z$ gives $$e^x(\cos y+i\sin y)=e^2(x+iy)$$.

Thus we have $$ \ \left\{ \begin{array}{ll} e^x\cos y=e^2x\\ \\ e^x\sin y =e^2 y \end{array} \right. \ $$

Manipulating this will give $e^{2x}=e^4(x^2+y^2)$ and $\tan y =\frac{y}{x}$. But I can neither give a contradiction nor prove there are roots given above formulas.

Considering this is a complex analysis question, maybe there are other ways of doing it instead of elementary ways?

Answer 1

Hint: try applying Rouché's Theorem to $f(z)=e^z-e^2z$ on the unit disc.