Re: complex analysis for trig. function

En el mensaje:1172279102.369717.320620@xxxxxxxxxxxxxxxxxxxxxxxxxxxx,
jennifer <scrilla_12_1999@xxxxxxxxx> escribió:
Hi am i trying to prove or disprove that: absolute value of sin z<= 1

where z is a complex number.

I tried looking at the definition of sin z in terms of the exponential
function i.e. (e^iz - e^-iz)/ 2i but i couldn't get anything from
this.

I only managed to show sin z <= 0 and i know this is true but i also
know that the absolute value of sin z is not necessary less than 1.

Can you help me please with a counter example or proof.

Thank you in advance.


If z = x + i*y, x and y real,

e^(i*z) = e^(i(x + i*y)) = e^(ix)*e^(-y)

e^(-i*z) = e^(-i(x + i*y)) = e^(-ix)*e^(y)

While |e^(+/-ix)| = 1, e^(+/-y) is unbounded.


--
Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@xxxxxxxxxxx


.

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