# complex analysis trigonometric inequalities 2

Using the definitions prove that $$|\sinh y| \le |\cos z|\le \cosh y \ , \ |\sinh y|\le|\sin z|\le \cosh y$$ Conclude that the complex cosine and sine are not bounded in the whole complex plane.

So I used the identity that $|\cos z|^2 = \cos^2(x) + \sinh^2(y)\ge |\sinh y|^2=\sinh^2(y)$. However, I'm not sure how to show that $|\cos z|^2 \le \cosh^2 y$.

Similarly for $|\sin z|$ I used the identity $|\sin z|^2 = \sin^2(x) + \sinh^2(y)\ge |\sinh y|^2$ and because we have the absolute value this implies $|\sinh y| \le |\sin z|$. However, again I am unsure how to show that $|\sin z| \le \cosh y$.

Also how do these inequalities allow us to conclude that the complex cosine and sine are not bounded in the whole complex plane?

• Hint: $\sin(z)=\sin(x+iy)=\sin(x)\cos(iy)+\sin(iy)\cos(x)=\sin(x)\cosh(y)+i\sinh(y) \cos(x)$. Can you show that $|\sin(z)|\le \cosh y$? For the conclusion think what you know about $|\sinh y|$. – Galc127 Aug 9 '16 at 5:42

We have that $$\cos(z)=\cos(x+iy)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y).$$ Notice that the above formula (or the inequality $|\sinh y| \le |\cos z|$) implies that along the imaginary axis the complex cosine is not bounded because $$|\cos(iy)|=|\cosh(y)|\to +\infty\quad\mbox{as y\to\pm\infty}.$$ As regards the inequality, by using the identity $\cosh^2(y)-\sinh^2(y)=1$ we get $$|\cos(z)|^2=[\cos(x)\cosh(y)]^2+[\sin(x)\sinh(y)]^2\\=\cos^2(x)\cosh^2(y)+\sin^2(x)[\cosh^2(y)-1]\\ =\cosh^2(y)-\sin^2(x)\leq \cosh^2(y).$$ Since $\cosh(y)\geq 1$ then we can conclude that $|\cos(z)|\leq \cosh(y)$.
As regards the complex sine $\sin(z)$, the approach is similar.