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I need to prove that $z=x+iy$ equals infinity is equivalent to $x = \infty$ and $y=\infty$.

I also have to give an example of a complex number $z$ so that $\sin(z)=\infty$.

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    Your first question amounts to showing the false if and only if false and false (which is easy). In other words, a complex number is _never_ infinity, nor are $x$ and $y$ in a valid expression $x+iy$ ever infinity. And your second question therefore has no good answer. \n\n Infinity only arises in taking limits of _expressions_, and so you should probably restate your question in those terms (however I don't think that statement is actually true in the usual sense of complex and real infinity, occuring as limits).2012-05-19

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The sine function is what's called an entire function, so $\sin z$ is finite for every complex number $z$.

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I'm guessing that some kind of a limit process is involved. But even so it is not clear how to "fix" the claim of the question to a true one. A few suggestions, mostly with a view of pointing out potential pitfalls to the OP.

If $t$ is a real parameter (if you prefer a sequence, then imagine that $t$ is a natural number), then $z(t)=e^t(\cos t +i \sin t)$ approaches $\infty$ in the sense of the definition that $|z|=e^t$ exceeds and stays above any predetermined bound. Yet neither the real $x(t)=e^t\cos t$ nor the imaginary $y(t)=e^t\sin t$ part of $z(t)$ tends to either (real) $+\infty$ or $-\infty$ due to the oscillating factor.

As Gerry said $\sin z$ is a finite complex number for all $z$. But $ \sin (it)=\frac{e^{i^2t}-e^{-i^2t}}{2i}=\frac{e^{-t}}{2i}-\frac{e^t}{2i} $ tends to the complex infinity, because the first term goes to zero, but the absolute value of the latter term $\to\infty$ as the real parameter $t\to\infty$. So you can say that $\sin z$ tends to $\infty$ when $z\to\infty$ along the imaginary axis.