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why isn't sinz bounded function? --anon Well, one has the equality
sin
(
z
)
=
1
2
i
(
e
i
z
−
e
−
i
z
)
{\displaystyle \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
(
1000
i
)
=
1
2
i
(
e
i
1000
i
−
e
−
i
1000
i
)
{\displaystyle \sin(1000i)={\frac {1}{2i}}\left(e^{i1000i}-e^{-i1000i}\right)}
=
1
2
i
(
e
−
1000
−
e
1000
)
.
{\displaystyle ={\frac {1}{2i}}\left(e^{-1000}-e^{1000}\right).}
Now,
e
−
1000
{\displaystyle e^{-1000}}
e
1000
{\displaystyle e^{1000}}
Oleg Alexandrov 21:05, 3 September 2005 (UTC) [ reply ]
how can one prove whether a function is bounded or not?
