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

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.

I am missing something trivial I know.

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

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 $\sinh y \geq 0$.

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Thank you, I knew I was missing something silly. – mccrack1985 Mar 20 '13 at 19:09
Prof also did it using exponential only. Thanks for the help. – mccrack1985 Mar 29 '13 at 10:58