Proving the Modulus of sin(z) is Greater or Equal to sin(x)

[shnaiwer]

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

 

[Integral]

What do you know about the modulus of complex numbers?

Note that on these forums we do not do proplems for you, we help, YOU do them.

 

[shnaiwer]

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)

 

[shnaiwer]

What do you know about the modulus of complex numbers?

Note that on these forums we do not do proplems for you, we help, YOU do them.

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

 

[Integral]

One proplem per thread please. Now what has your first post got to do with sin(z)?

 

[shnaiwer]

One proplem per thread please. Now what has your first post got to do with sin(z)?

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 ?

 

[Office_Shredder]

Your original post is asking about the modulus of a complex number; no question here involves sin(z). Unless you specify what you're trying to answer we can't help

 

[shnaiwer]

Your original post is asking about the modulus of a complex number; no question here involves sin(z). Unless you specify what you're trying to answer we can't help

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

thank u

 

[shnaiwer]

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

 

[shnaiwer]

where r uuuuuuuuuuu

 

[Dick]

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

That much is true. The rest isn't. Find the square of the modulus of sin(z) by multiplying sin(z) by it's complex conjugate.

 

[shnaiwer]

That much is true. The rest isn't. Find the square of the modulus of sin(z) by multiplying sin(z) by it's complex conjugate.

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 ) 2)
= ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 - ( sinh y)2 ( sin x ) 2
= ( sin x ) 2 (( cosh y ) 2 - ( sinh y)2 ) + ( sinh y)2
so , we have

│sin z │2 = ( sin x ) 2 + ( sinh y)2

but the problem of y variable still hold !

can i say that :
( sin x ) 2 + ( sinh y)2 >= ( sin x ) 2 ?

 

[Dick]

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 ) 2)
= ( sin x ) 2 ( cosh y ) 2 + ( sinh y)2 - ( sinh y)2 ( sin x ) 2
= ( sin x ) 2 (( cosh y ) 2 - ( sinh y)2 ) + ( sinh y)2
so , we have

│sin z │2 = ( sin x ) 2 + ( sinh y)2

but the problem of y variable still hold !

can i say that :
( sin x ) 2 + ( sinh y)2 >= ( sin x ) 2 ?

Sure you can. sinh(y)^2>=0. Just because it's the square of something.

 

[shnaiwer]

Sure you can. sinh(y)^2>=0. Just because it's the square of something.

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 ?

 

[Dick]

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 ?

That's fine. But you don't sound very sure of yourself?

 

[shnaiwer]

That's fine. But you don't sound very sure of yourself?

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