Show that $\cos(z)=z$ has a solution for some complex number $z$ (non-real)

Question

I've tried using the exponential form of cosine but the equation was harder than it seemed to solve. I've seen a solution on this site using a Picard Theorem but I don't have any knowledge of analysis yet (I am in high school). Is there a way to show this using elementary maths?

Answer 1

Using $z=x+i y$, expanding to cosine and isolating the real and imaginary parts, we have the two equations $$\cos (x) \cosh (y)-x=0 \tag 1$$ $$\sin (x) \sinh (y)+y=0 \tag 2$$

The first equation gives $$y=\cosh ^{-1}(x \sec (x))\tag 3$$ which, inserted in the second, would give (assuming $x>1$) $$\tan (x)\sqrt{x^2-\cos ^2(x)} +\cosh ^{-1}(x \sec (x))=0\tag 4$$ which is impossible to solve explicitly even using special functions.

Graphing, what you could notice is that the $n^{\text{th}}$ zero of function $$f(x)=\tan (x)\sqrt{x^2-\cos ^2(x)} +\cosh ^{-1}(x \sec (x))$$ are closer and closer to $2n\pi$. Using series expansion $$f(x)=\cosh ^{-1}(2 \pi n)+\frac{4 \pi ^2 n^2}{\sqrt{4 \pi ^2 n^2-1}} (x-2 \pi n)+O\left((x-2 \pi n)^2\right)$$ giving, as an approximation, $$x_n=2n\pi-\frac{\sqrt{4 \pi ^2 n^2-1} }{4 \pi ^2 n^2}\cosh ^{-1}(2 \pi n)\tag 5$$ which, expanded again for large $n$, gives $$x_n=2 \pi n-\frac{\log (4 \pi n)}{2 \pi n}+O\left(\frac{1}{n^3}\right)$$ leading to $$y_n=\log (4 \pi n)+O\left(\frac{1}{n^2}\right)$$ Using $(3)$ and $(5)$ seems to give decent approximations $$\left( \begin{array}{cccc} n & x_{\text{estimate}}& y_{\text{estimate}}& x_{\text{solution}}& y_{\text{solution}}\\ 1 & 5.8864994 & 2.5404479 & 5.8695604& 2.5448858 \\ 2 & 12.310739 & 3.2351074 & 12.308558& 3.2355041 \\ 3 & 18.657306 & 3.6372873 & 18.656748& 3.6373661 \\ 4 & 24.977015 & 3.9228865 & 24.976832& 3.9229079 \\ 5 & 31.284206 & 4.1447097 & 31.284142& 4.1447162 \\ 6 & 37.584491 & 4.3261473 & 37.584472& 4.3261491 \\ 7 & 43.880537 & 4.4796761 & 43.880536& 4.4796762 \\ 8 & 50.173780 & 4.6127518 & 50.173787& 4.6127514 \\ 9 & 56.465068 & 4.7301900 & 56.465078& 4.7301894 \\ 10 & 62.754935 & 4.8352824 & 62.754945& 4.8352818\\ 20 & 125.61973 & 5.5273581 & 125.61973& 5.5273579 \\ 30 & 188.46409 & 5.9325429 & 188.46409& 5.9325429 \\ \end{array} \right)$$

Edit

We can do better in terms of approximations expanding $f(x)$ as a series around $x=2n\pi$ and continuing with power series reversion.

To make the results more compact, let $$\color{blue}{m=2n\pi} \qquad \text{and}\qquad \color{blue}{t=\frac{\sqrt{m^2-1} }{m^2}\,\cosh ^{-1}(m)}$$ This gives $$\color{red}{x=m-t+\sum_{k=2}^\infty \frac{P_k}{Q_k}\,t^k}$$ where the first polynomials in $m$ are $$\left( \begin{array}{cc} k & P_k \\ 2 & 3 m^2-4 \\ 3 & m^4-14 m^2+20 \\ 4 & 25 m^6-240 m^4+528 m^2-320 \\ 5 & -3 m^8+179 m^6-1324 m^4+2816 m^2-1760 \\ 6 & -441 m^{10}+12537 m^8-86520 m^6+227376 m^4-251328 m^2+98560 \\ \end{array} \right)$$ and $$\left( \begin{array}{cc} k & Q_k \\ 2 & -2 m\left(m^2-1\right) \\ 3 & 3 m^2 \left(m^2-1\right) \\ 4 & 12 m^3 \left(m^2-1\right)^2 \\ 5 & 15 m^4 \left(m^2-1\right)^2 \\ 6 & 180 m^5 \left(m^2-1\right)^3 \\ \end{array} \right)$$

which leads to much better estimates. $$\left( \begin{array}{ccc} n & x_{\text{estimate}}& x_{\text{solution}} \\ 1 & 5.8698165 & 5.8695604 \\ 2 & 12.308553 & 12.308557 \\ 3 & 18.656747 & 18.656748 \\ 4 & 24.976832 & 24.976832 \\ \end{array} \right)$$