Show that $|\cos z| \leq e$ if $|z|=1$

Question

I'm working on a homework problem, and a part of it is to show that $|\cos z| \leq e$ if $|z|=1$, where $z$ is a complex number.

I figured that it would be useful to write $\cos(z) = \frac{e^{iz}+e^{-iz}}{2}$, and then by the triangle inequality, we can say that $|\cos(z)| \leq 1/2 |e^{iz}| + 1/2 |e^{-iz}|$, however working out the two norms yields $1/2(2)=1$, which isn't very useful.

Any hints would be greatly appreciated.

Answer 1

Using the definition: writing $z=a+ib$ for $a,b\in\mathbb R$, $$ |\cos z|=\left|\frac{e^{iz}+e^{-iz}}2\right|=\left|\frac{e^{ia}e^{-b}+e^{-ia}e^{b}}2 \right| \leq\frac{|e^{ia}|\,e+|e^{-ia}|\,e}{2}=e, $$ where we have used that $|b|\leq1$.

Answer 2

If $|z|=1$, then $z=\cos t+i\sin t$ for some $t$ real, hence $$2\cos z=\mathrm e^{iz}+=\mathrm e^{-iz}=\mathrm e^{-\sin t}\mathrm e^{i\cos t}+\mathrm e^{\sin t}\mathrm e^{-i\cos t}.$$ One knows that $|\mathrm e^{i\cos t}|=|\mathrm e^{-i\cos t}|=1$ hence, by the triangular inequality, $$2|\cos z|\leqslant\mathrm e^{-\sin t}+\mathrm e^{\sin t}.$$ The function $u\mapsto\mathrm e^{-u}+\mathrm e^{u}$ defined on $[-1,1]$ is maximal at $u=\pm1$ hence $$2|\cos z|\leqslant\mathrm e^{-1}+\mathrm e,$$ which is a refinement of the result to prove.

Answer 3

This is (more or less) just a rephrasing of Did's answer, but since: $$\cos z = \sum_{j=0}^{+\infty}\frac{(-1)^j z^{2j}}{(2j)!}$$ we have that $|z|=1$ implies: $$|\cos z|\leq \sum_{j=0}^{+\infty}\frac{1}{(2j)!}=\cosh 1<e.$$

Answer 4

If $\vert z \vert = 1$, then we may write

$z = \cos \theta + i\sin \theta \tag{1}$

for $\theta \in [0, 2\pi).$ Then

$iz = -\sin \theta + i\cos \theta, \tag{2}$

so that

$e^{iz} = e^{-\sin \theta + i\cos \theta} = e^{-\sin \theta} e^{i\cos \theta}, \tag{3}$

from which we have

$\vert e^{iz} \vert = \vert e^{-\sin \theta} e^{i\cos \theta} \vert = \vert e^{-\sin \theta} \vert \vert e^{i\cos \theta} \vert = e^{-\sin \theta}, \tag{4}$

since $e^{-\sin \theta}$ is positive real and $\vert e^{i\cos \theta} \vert = 1$. A similar calculation reveals that

$\vert e^{-iz} \vert = e^{\sin \theta}. \tag{5}$

Using (4) and (5) in the inequality

$\vert \cos(z) \vert \leq 1/2 \vert e^{iz} \vert + 1/2 \vert e^{-iz} \vert \tag{6}$

immediately yields

$\vert \cos(z) \vert \leq \dfrac{1}{2}(e^{-\sin \theta} + e^{\sin \theta}) \le e, \tag{7}$

since $\vert \pm\sin \theta \vert \le 1$ for all $\theta$. Notice that $(7)$ actually yields the tighter bound,

$\vert \cos(z) \vert \le \dfrac{1}{2}(e + e^{-1}), \tag{8}$

as observed and deomonstrated by Did and Jack D'Aurizio in their answers.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!