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let
then
and
so
Hence:
We need to show that this is always between -1 and 1 inclusive, that is to say we need to show that the denominator always has greater or equal absolute value than that of the numerator, i.e. that:
i.e. that
i.e. that
i.e. that
i.e. that
Which is clearly true as the LHS is always > 0.
One thing, we have proven the inequality for ALL complex numbers except one, the number 0 + 0i, for this number the inequality is not true as the denominator would be zero.
Last edited by gnitsuk (2011-11-24 23:55:53)
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