Finding Solutions to the Sinh Equation in Complex Numbers

[Benny]

Hello, can someone please help me out with the following question?

Express [tex]\sinh \left( {x + yi} \right)[/tex] in terms of [tex]\sinh (x)[/tex], [tex]\cosh(x)[/tex], [tex]\cos(y)[/tex] and [tex]\sin(y)[/tex]. Hence find all solutions [tex]z \in C[/tex] of the equation [tex]\sinh(z) = i[/tex].

I have little idea as to how to do this question. Here is my working.

[tex]\sinh \left( {x + yi} \right) = \sinh \left( x \right)\cosh \left( {iy} \right) + \cosh \left( x \right)\sinh \left( {iy} \right)[/tex]

[tex] = \sinh \left( x \right)\left( {\frac{{e^{iy} + e^{ - iy} }}{2}} \right) + \cosh \left( x \right)\left( {\frac{{e^{iy} - e^{- iy} }}{2}} \right)[/tex]

[tex]{\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)[/tex]

[tex] = \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right)[/tex]

So [tex]\sinh \left( z \right) = i \Rightarrow \sinh \left( x \right)\cos \left( y \right) + i\cosh \left( x \right)\sin \left( y \right) = i[/tex]

Equating real and imaginary parts:

[tex]\sinh \left( x \right)\cos \left( y \right) = 0...(1)[/tex]

[tex]\cosh \left( x \right)\sin \left( y \right) = 1...(2)[/tex]

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.

[tex]\sinh \left( x \right)\cos \left( y \right) = 0[/tex]

[tex] \Rightarrow \cos (y) = 0[/tex] or [tex]\sinh (x) = 0[/tex]

[tex]\frac{{e^x - e^{ - x} }}{2} = 0 \Rightarrow x = 0[/tex] and [tex]y = \frac{{n\pi }}{2}[/tex] 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 [tex]\sin \left( y \right) = 1[/tex] which has solutions: [tex]y = \frac{\pi }{2} + 2k\pi[/tex] 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 [tex]y = \frac{\pi }{2} + 2k\pi[/tex].

So I end up with [tex]\sinh \left( z \right) = i \Leftrightarrow z = \left( {\frac{\pi }{2} + 2k\pi } \right)i[/tex]. 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?

 

[dextercioby]

It looks okay to me...

You may have tried to check whether

[tex] \sinh\left(\frac{i\pi}{2}+2ki\pi\right)=i [/tex]


Daniel.

 

[Benny]

Thanks for your response dextercioby. I have checked that my solution works for k = 0 but not for any other values.