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Find a sequence of complex numbers ${z_n}$ such that $\sin z_n$ is real for all $n$ and tends to $\infty$ as $n→\infty$ ?

I get an example as $\log 2in$ . I want to verify that am I right or wrong. If I am right then can anyone give me some more examples.

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Integral deleted his/her comment, but the solution was a good one: it is just that the sequence given worked for cosine, not sine. But that is easily rectified, and the sequence $z_n=\frac{\pi}{2}+in$ works just fine. –  user641 Mar 13 '13 at 15:39
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2 Answers 2

$z_n=\arcsin a_n$ where $a_n$ is any real-valued sequence which tends to $\infty$.

The inverse of the sine can be calculated via: $$\arcsin w=\frac{1}{i} \log \left( i w+\sqrt{1-w^2} \right)$$

(other branches are possible, of course)

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So to give a specific example, $z_n=\pi/2 + \frac{1}{i}\log(n+\sqrt{n^2-1})$. –  user641 Mar 13 '13 at 15:35
can anyone tell me please that my answer is right or not? –  user65544 Mar 13 '13 at 15:37
@user65544: No, your answer is not right. –  user641 Mar 13 '13 at 15:38
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You can do this systematically: let $z = x + i y$, and you find $$ \sin z = \frac{1}{2i}\left( e^{-y}(\cos x + i \sin x) - e^{-y}(\cos x - i\sin x) \right) $$

Next you need the imaginary parts to cancel. Then you'll find an expression for the real part in terms of some parameters, which you can choose appropriately to get a sequence tending to $\infty$.

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