Talk:Bounded function

Sin z not bounded?

The function f:R → R defined by f (x)=sin x is bounded. The sine function is no longer :bounded if it is defined over the set of all complex numbers.

why isn't sinz bounded function? --anon

Well, one has the equality

$\sin(z)={\frac {1}{2i}}\left(e^{iz}-e^{-iz}\right)$

which follows from Euler's formula. If z=1000i, which is an imaginary number, one has

$\sin(1000i)={\frac {1}{2i}}\left(e^{i1000i}-e^{-i1000i}\right)$

$={\frac {1}{2i}}\left(e^{-1000}-e^{1000}\right).$

Now, $e^{-1000}$ is small, but $e^{1000}$ is huge, so this adds up to a huge number. Does that make sence? Oleg Alexandrov 21:05, 3 September 2005 (UTC)