# Show that sin z

1. Mar 9, 2009

### 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

Last edited: Mar 9, 2009
2. Mar 9, 2009

### Integral

Staff Emeritus
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.

3. Mar 9, 2009

### shnaiwer

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)

4. Mar 9, 2009

### shnaiwer

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

5. Mar 9, 2009

### Integral

Staff Emeritus

6. Mar 9, 2009

### shnaiwer

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 ?

7. Mar 9, 2009

### Office_Shredder

Staff Emeritus
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

8. Mar 9, 2009

### shnaiwer

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

thank u

9. Mar 11, 2009

### 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

10. Mar 11, 2009

### shnaiwer

where r uuuuuuuuuuu

11. Mar 11, 2009

### Dick

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.

12. Mar 13, 2009

### shnaiwer

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 ??????

13. Mar 13, 2009

### Dick

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

14. Mar 13, 2009

### shnaiwer

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 ???

15. Mar 13, 2009

### Dick

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

16. Mar 13, 2009

### shnaiwer

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