- #1
Benny
- 577
- 0
Hello, can someone please help me out with the following question?
Express \sinh \left( {x + yi} \right) in terms of \sinh (x), \cosh(x), \cos(y) and \sin(y). Hence find all solutions z \in C of the equation \sinh(z) = i.
I have little idea as to how to do this question. Here is my working.
\sinh \left( {x + yi} \right) = \sinh \left( x \right)\cosh \left( {iy} \right) + \cosh \left( x \right)\sinh \left( {iy} \right)
= \sinh \left( x \right)\left( {\frac{{e^{iy} + e^{ - iy} }}{2}} \right) + \cosh \left( x \right)\left( {\frac{{e^{iy} - e^{- iy} }}{2}} \right)
{\rm = sinh}\left( {\rm x} \right)\left( {\frac{{e^{iy} + e^{iy} }}{2}} \right) + i\cosh \left( x \right)\left( {\frac{{e^{iy} - e^{ - iy} }}{{2i}}} \right)
= \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right)
So \sinh \left( z \right) = i \Rightarrow \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right) = i
Equating real and imaginary parts:
\sinh \left( x \right)\cos \left( y \right) = 0...(1)
\cosh \left( x \right)\sin \left( y \right) = 1...(2)
At this point I am unsure of how to proceed. The equations do not look solvable, directly anyway. So I decided to start off with equation 1 and see where that would lead. Here is what I have done.
\sinh \left( x \right)\cos \left( y \right) = 0
\Rightarrow \cos (y) = 0 or \sinh (x) = 0
\frac{{e^x - e^{ - x} }}{2} = 0 \Rightarrow x = 0 and y = \frac{{n\pi }}{2} where n is an odd ineger.
Now since equations (1) and (2) need to be satisfied simultaneously then x is necessarily equal to zero and equation (2) reduces to \sin \left( y \right) = 1 which has solutions: y = \frac{\pi }{2} + 2k\pi where k is an integer.
I need y to satisfy both equations (1) and (2) so I take y to be the 'least general' answer so that y = \frac{\pi }{2} + 2k\pi.
So I end up with \sinh \left( z \right) = i \Leftrightarrow z = \left( {\frac{\pi }{2} + 2k\pi } \right)i. I'm not sure if my method is right. I pretty much got stuck at the simultaneous equations part. Can someone please hep me out with this question?
Express \sinh \left( {x + yi} \right) in terms of \sinh (x), \cosh(x), \cos(y) and \sin(y). Hence find all solutions z \in C of the equation \sinh(z) = i.
I have little idea as to how to do this question. Here is my working.
\sinh \left( {x + yi} \right) = \sinh \left( x \right)\cosh \left( {iy} \right) + \cosh \left( x \right)\sinh \left( {iy} \right)
= \sinh \left( x \right)\left( {\frac{{e^{iy} + e^{ - iy} }}{2}} \right) + \cosh \left( x \right)\left( {\frac{{e^{iy} - e^{- iy} }}{2}} \right)
{\rm = sinh}\left( {\rm x} \right)\left( {\frac{{e^{iy} + e^{iy} }}{2}} \right) + i\cosh \left( x \right)\left( {\frac{{e^{iy} - e^{ - iy} }}{{2i}}} \right)
= \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right)
So \sinh \left( z \right) = i \Rightarrow \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right) = i
Equating real and imaginary parts:
\sinh \left( x \right)\cos \left( y \right) = 0...(1)
\cosh \left( x \right)\sin \left( y \right) = 1...(2)
At this point I am unsure of how to proceed. The equations do not look solvable, directly anyway. So I decided to start off with equation 1 and see where that would lead. Here is what I have done.
\sinh \left( x \right)\cos \left( y \right) = 0
\Rightarrow \cos (y) = 0 or \sinh (x) = 0
\frac{{e^x - e^{ - x} }}{2} = 0 \Rightarrow x = 0 and y = \frac{{n\pi }}{2} where n is an odd ineger.
Now since equations (1) and (2) need to be satisfied simultaneously then x is necessarily equal to zero and equation (2) reduces to \sin \left( y \right) = 1 which has solutions: y = \frac{\pi }{2} + 2k\pi where k is an integer.
I need y to satisfy both equations (1) and (2) so I take y to be the 'least general' answer so that y = \frac{\pi }{2} + 2k\pi.
So I end up with \sinh \left( z \right) = i \Leftrightarrow z = \left( {\frac{\pi }{2} + 2k\pi } \right)i. I'm not sure if my method is right. I pretty much got stuck at the simultaneous equations part. Can someone please hep me out with this question?
