# Suppose that cos z is a real number of absolute value greater than 1. What is z?

Suppose that cos z is a real number of absolute value greater than 1. What is z? I am really puzzled.

• Please tell us a little bit about your level of familiarity with complex functions, and what approaches you've tried with this question. For example, have you seen how to decompose $\cos z$ into real and imaginary parts? – Jonathan Y. Oct 25 '13 at 14:29
• Hi thx for the reply. I have done one semester of complex analysis and this is the homework of another class. I have tried using the formula $\frac{e^{zi}+e^{-zi}}{2}$ and take $w=e^{zi}$. However I ended up with a very complicated expression that I could not solve. – user94602 Oct 25 '13 at 14:46
• Well, @RobertIsrael pretty much pointed you in the same direction, although without proving that the known identity of $\cos(\alpha+\beta)$ for reals holds for imaginary angles, it might seem a little mysterious. – Jonathan Y. Oct 25 '13 at 14:48
• Oh..Thank you very much! – user94602 Oct 25 '13 at 14:50
• sorry i still can not get an elegant result that looks right – user94602 Oct 25 '13 at 15:08

This is just a little too long for a comment, but is intended to expand the comments above: $$\cos z = \cos(x+iy) = \frac{1}{2}\left(e^{i(x+iy)}+e^{-i(x+iy)}\right) = \frac{1}{2}\left(e^{-y}(\cos x+i\sin x) + e^{y}(\cos x -i\sin x)\right) = \cos x\frac{e^y+e^{-y}}{2} + \sin x\frac{i(e^{-y}-e^y)}{2} = \cos x\cosh y - i\sin x\sinh y$$ This should help in figuring out when $\Im\cos z=0$ and what that implies for $\Re\cos z$.
• so $(sinxsinhy)^2+(cosxcoshy)^2 > 1$? – user94602 Oct 25 '13 at 16:00
Hint: Expand $\cos(x+iy)$ using the sum formula for $\cos$.
• $\cos(x+iy) = \cos(x)\cos(iy) - \sin(x) \sin(iy) = \cos(x) \cosh(y) - i \sin(x) \sinh(y)$ – Robert Israel Oct 27 '13 at 2:00