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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). –  Marc van Leeuwen May 19 '12 at 10:37
    
thank you very much. –  johnny May 19 '12 at 10:41

2 Answers 2

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.

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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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