Finding Solutions to the Sinh Equation in Complex Numbers

[Benny]

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?

 

[dextercioby]

It looks okay to me...

You may have tried to check whether

$$\sinh\left(\frac{i\pi}{2}+2ki\pi\right)=i$$


Daniel.

 

[Benny]

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