# Prove that the equation has $n$ roots in the unit disk

If $k＞1$, prove $z^n\exp(k-z)＝1$ has $n$ roots in unit disk $Δ(0,1)$.

I want to use Rouche's theorem. But I don't know how to choose another function to compare.

• Hint. You should use the principle of argument instead. Jan 3, 2018 at 12:14
• @CaveJohnson thanks,then it could be calculated... Jan 3, 2018 at 12:46

Transform it to $$z^n-\exp(z-k)=0$$ and apply Rouché with the exponential term as perturbation.
$$\frac{1}{2\pi i}\oint_{|z|=1}\frac{z^n e^{k-z}\left(\frac{n}{z}-1\right)}{z^n e^{k-z}-1}\,dz$$ counts the number of zeroes of $z^n e^{k-z}-1$ in the unit disk (there are no poles, of course, and it is not difficult to check that there are no zeroes on the boundary of the unit disk). The previous integral can be written as $$\frac{1}{2\pi i}\oint_{|z|=1}\frac{\left(\frac{n}{z}-1\right)}{1-e^{z-k} z^n}\,dz$$ and since the modulus of $e^{z-k}z^n$ over $|z|=1$ is strictly less than one, such integral can be written as $$\frac{1}{2\pi i}\oint_{|z|=1}\left(\frac{n}{z}-1\right)\,dz+\frac{1}{2\pi i}\sum_{m\geq 1}\oint_{|z|=1}\left(\frac{n}{z}-1\right)e^{mz-mk}z^{mn}\,dz$$ where the first term equals $n$ and the second term equals zero, since it is the integral of a holomorphic function over a closed contour.