Recent content by shnaiwer

thank u ... " | " denote nothing other than ( when or at ) , we can neglect it.. when we sub in zero we get zero , since ( 0^z) * e^(-0) = 0 * 1 = 0 but the problem still hold when we try that limit as x goes to infinity , where it gives ( ∞ / ∞ ) or ( 0 / 0) if we try l'Hôpital's...

hi to all ... it is just an attempt to understand ... the problem was to prove that gamma ( z+ 1) = z * gamma(z) using the integral definition of gamma function ... when i used the integration by parts i get the following : [SIZE="5"][SIZE="5"]gamma ( z+ 1 ) = ( x^z) * e^(-x) (0...

may be ... thank you very much ...

so , then i can say that │sin z │2 >= ( sin x ) 2 taking square root to both sides gives │sin z │>= │sin x│ did it finish ?

thanx ... a lot i tried and wrote the following .. Sin ( z ) = sin ( x + iy ) = sin x cos iy + sin iy cos x = sin x cosh y + i sinh y cos x Now │sin z │2 = ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 (cos x )2 = ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 (1 – ( sin x...

where r uuuuuuuuuuu

no one will try with me ! ok i say : sin z = sin ( x + iy ) = sin x cos iy + cos x sin iy = sin x cosh y - i cos x sinh y modulus (sin z) > = modulus ( sin x cosh y) - modulus ( i cos x sinh y) is this trueeeeeeeeee

agin i am sorry ... i mean sin z and sin x thank u

really i am sorry .. sin z = ( exp i(x+iy) - exp -i(x+iy))/2i ( sin z )* = exp ( -i(x-iy) - exp ( i(x-iy)/(-2i) (mod. (sin z) )^2 = ( exp i(x+iy) - exp -i(x+iy))/2i * exp ( -i(x-iy) - exp ( i(x-iy)/(-2i) is this step good ?

i can say that (modulus fo sin z )^2 = sin z * (sin z )*

also i am asking about this problem : what part of z - plane corresponds to the interior of the unit circle in the w-plane if a) -w = (z-1)/(z+1) b) -w = (z - i)/(z + i)

hi to all can i obtain the answer of this problem : show that : modulus of (sin z ) > = modulus of (sin X ) where z = x + i Y