How do I prove that a complex number equals infinity? 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 $\sin(z)=\infty$.
 A: The sine function is what's called an entire function, so $\sin z$ is finite for every complex number $z$. 
A: 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.
