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Let $t,x \in \mathbb{R}$. I need to show that $\lim_{t \to \pm \infty} \ |\cos(x+it)| = \infty$ and since I didn't do Calculus for quite some time I wanted to ask whether my solution is correct. (I only used some basic facts about hyperbolic functions.):

$ \lim_{t \to \pm \infty} \ |\cos(x+it)| $

$ = \lim_{t \to \pm \infty}|\cos(x)\cos(it)-\sin(x)\sin(it)|$

$ = \lim_{t \to \pm \infty}|\cos(x)\cosh(t)-\sin(x)\sinh(t)/(-i)|$

$ = \lim_{t \to \pm \infty}|\cos(x)\cosh(t)+i\sin(x)\sinh(t)|$

By definition of the absolute value in $\mathbb{C}$:

$ = \lim_{t \to \pm \infty} \cos^2(x)\cosh^2(t)-\sin^2(x)\sinh^2(t)$

By use of elementary identities:

$ = \lim_{t \to \pm \infty}[1-\sin^2(x)][1-\sinh^2(t)] - \sin^2(x)\sinh^2(t)$

$ = \lim_{t \to \pm \infty} 1-\sinh^2(t) - \sin^2(x) + \sin^2(x)\sinh^2(t) - \sin^2(x)\sinh^2(t)$

$ = \lim_{t \to \pm \infty} \ 1-\sinh^2(t) - \sin^2(x)$

$ = \lim_{t \to \pm \infty} \cosh^2(t) - \sin^2(x)$

Since $\lim_{t \to \pm \infty} \cosh(t) = \infty$:

$ = \infty$

Thanks for any answers.

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You're making it much harder than it needs to be. Just rewrite:

$$\cos(x+ i t) == \frac{1}{2}\left(e^{i x-t} + e^{-i x + t}\right)$$

Then, clearly as $t \to \infty$ the norm of the expression diverges.

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