Prove $|Im(z)|\le |\cos (z)|$ for $z\in \mathbb{C}$.

Question

Let $z\in \mathbb{C}$ i.e. $z=x+iy$. Show that $|Im(z)|\le |\cos (z)|$.

My hand wavy hint was to consider $\cos (z)=\cos (x+iy)=\cos (x)\cosh (y)+i\sin (x)\sinh(y)$ then do "stuff".

Then I have $|\cos (z)|=|Re(z)+iIm(z)|$ and the result will be obvious.

Thanks in advance.

I am missing something trivial I know.

Answer 1

$$\cos(z) = \cos(x) \cosh(y) + i \sin(x) \sinh(y) $$

Taking the norm squared: $$ \cos^2(x) \cosh^2(y) + \sin^2(x) \sinh^2(y)$$

We are left with:

$$ \cos^2(x) \left(\frac{1}{2}\cosh(2y) + \frac{1}{2}\right) + \sin^2(x) \left(\frac{1}{2}\cosh(2y) - \frac{1}{2}\right)$$ Simplifying, we get: $$ \frac{1}{2} \left(\cosh(2y) + \cos(2x) \right)$$

We might as well suppose $\cos{2x} = -1$ Our goal is to show $|$Im$(z)|^2 $ is smaller than this quantity.

That is,

$$ \begin{align} & & y^2 & \leq \frac{1}{2} \cosh{2y} - \frac{1}{2} \\ \iff & & y^2 &\leq (\sinh y)^2 \\ \iff & & y & \leq \sinh y \ \ \ \ \ \ \ \ \forall y\geq0\end{align}$$

A quick computation of the derivative shows that $\frac{d}{dy} y = 1$ but $\frac{d}{dy} \sinh y = \cosh{y}$. If we want to see that $\cosh y \geq 1$, we can differentiate it again and see that $\sinh y \geq 0$.