Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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

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 $\sinh y \geq 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.