Solving $z^3 e^{1-z}=1$ inside the unit circle

Question

Prove that $z^3 e^{1-z}=1$ has exactly two roots inside the circle $|z|=1$. I showed that $z=1$ is a solution on the boundary of the circle, how can I find the other solution?

Answer 1

We have $f(z)=z^3e^{1-z}-1$ and $\gamma=re^{it}$, where $r\gt 0$ and $0\leq t\leq 2\pi$. The goal is to find the total number of zeros of $f$ inside the closed contour $\gamma$ when $r=1$, also known as the unit disk. Note that $f$ does have a zero at $z=1$, however this zero is on the contour $\gamma$ when $r=1$ and not inside it. In order to use the tools that follow, we must use $\gamma$ when $r\lt 1$ but very close to $1$, such as $0.8$ or $0.9$, because $f$ cannot have zeros on the contour.

Argument Principle

Since $f$ is indeed a meromorphic function on and inside a closed contour $\gamma$ and $f$ has no zeros or poles on $\gamma$, then we can utilize the argument principle. Which states that $$\frac{1}{2\pi i}\oint_\gamma\frac{\frac{\mathrm d}{\mathrm dz}f(z)}{f(z)}\ \mathrm dz=Z_{f,\gamma}-P_{f,\gamma}$$ Where $Z_{f,\gamma}$ is the sum of the multiplicity of each zero of $f$ inside $\gamma$ and $P_{f,\gamma}$ is the sum of the order of each pole of $f$ inside $\gamma$. In this case, it's clear to see that $f$ has no poles, therefore $P_{f,\gamma}=0$.

Winding Number

By using the substitution $w=f(z)$, we have $$\frac{1}{2\pi i}\oint_\gamma\frac{\frac{\mathrm d}{\mathrm dz}f(z)}{f(z)}\ \mathrm dz=\frac{1}{2\pi i}\oint_{f\circ\gamma}\frac{1}{w}\ \mathrm dw$$ Where the contour integral on the right-hand side is the winding number of the curve $f\circ\gamma$ about the origin.

Curve Mapping

At this point, we're left with $$\frac{1}{2\pi i}\oint_{f\circ\gamma}\frac{1}{w}\ \mathrm dw=Z_{f,\gamma}$$ Where $$f\circ\gamma=f(\gamma(t))=r^3e^{3it}e^{1-re^{it}}-1$$ In the image below, we have $\gamma$ in red, the mapped curve $f\circ\gamma$ in blue and three points which are zeros. Note that in this image, $r=0.8$. At this point it should be clear that $f\circ\gamma$ winds around the origin twice in a counterclockwise direction. Which implies that $$\frac{1}{2\pi i}\oint_{f\circ\gamma}\frac{1}{w}\ \mathrm dw=2$$ Therefore, $f$ contains $2$ zeros inside the unit disk.