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Prove that the equation $z^{3}e^{z}=1$ has infinitely many complex solutions.How many of them are real?

Use the argument principle,I choose a disk centered at $0$ with radius $R$ and get $\int_{\partial D}\dfrac {3z^{2}e^{z}+z^{3}e^{z}}{z^{3}e^{z}-1}dz$,and I don't how to do with this integral.Additionally,I use the monotonicity when z $z$ takes value in real number to find that there could be one real solution.

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  • $\begingroup$ Use argument principle can solve it but there is a more succinct way. $\endgroup$
    – Zack Ni
    Jul 27, 2016 at 3:16
  • $\begingroup$ @ZackNi I don't know how to use argument to solve this problem.Can you give me some hint. $\endgroup$
    – Jack
    Jul 29, 2016 at 1:50

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Set $f(z)=z^3e^z-1$, and let $g(z)$ be the meromorphic function defined by $g(z)=f(1/z)$. It is easy to check, by looking at a power series expansion, that $g$ has an essential singularity at $0$. So by Picard’s Theorem, for every neighborhood $U$ of $0$, there exists $w \in \mathbb{C}$ so that $g(U \setminus \{0\}) = \mathbb{C} \setminus \{w\}$. Now, for each $n > 0$, set $U_n = \{0 < |z| < 1/n\}$. We know $-1 \notin g(U_n)$, since $g(z)=-1$ implies $f(1/z)+1=0$ and thus $1/z=0$, an impossibility. So we must have $g(U_n) = \mathbb{C} \setminus \{-1\}$. In particular, there exists $z_n$ with $g(z_n) = 0$. We then have $\{1/z_n\}_{n \in \mathbb{N}}$ providing a set of solutions, and it is necessarily infinite, since $|1/z_n| > n$ for each $n$. (I apologize if Picard’s Theorem is too heavy-duty for what you’re looking for. I’m not sure how to do it any other way.)

Now consider when $z=x \in \mathbb{R}$. If $x \leq 0$, then $x^3e^x \leq 0$, so $f(x)$ is negative. On the other hand, $f'(x) = (x^3+3x^2)e^x$, so $f'(x) > 0$ for all $x > 0$. This means $f$ crosses the $x$-axis at most once. And, indeed, it crosses at least once, since $f(0)=-1$ and $f(1)>0$. So $f$ has a zero in $(0,\infty)$, hence one real zero overall.

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Just for your curiosity.

The equation $z^3\,e^z=1$ has three roots which express in terms of Lambert function. They are $$x_1=3 W\left(\frac{1}{3}\right)$$ $$x_2=3 W\left(-\frac{1}{3} (-1)^{1/3}\right)$$ $$x_3=3 W\left(\frac{1}{3} (-1)^{2/3}\right)$$ Since the argument is complex for $x_2$ and $x_3$, there is only one real root which is $x_1$.

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