Question : What are real and imaginary parts of tanh z ?
Solution :
To find real and imaginary parts of tanh z where z is a complex number.
A complex number z is written as
\[z=x+iy\]
where x and y are real numbers.
We know that
\[tan iz = i tanh z\]
\[tanh z =frac{1}{i} tan iz = – i tan iz\]
for z =x+iy => iz =ix -y
\[tanh (x+iy) = – i tan (ix-y)\]
Let A and B are real and imaginary parts of tanh z.
\[=>A+iB = tanh z = – i tan (ix-y)\]
then it’s conjugate
\[=>A -iB = -i tan (ix+y) \]
adding these two equations
\[=> 2A =- i tan (ix-y)-i tan (ix+y)\]
\[ => A= -i \frac{sin 2ix}{2cos(ix-y)(cos(ix-y)} \]
2cos A cos B =Cos(A+B)+Cos(A-B)
\[A=\frac{-i}{2}\frac{sin 2ix}{cos(2ix)+(cos(2y)}\]
\[A=\frac{sinh 2x}{cosh (2x)+(cos(2y)}\]
This is required real part if tanh z
To get imaginary part of tanh z, subtracting equations for A+iB and A-iB
\[2iB = \frac{1}{i} tan(ix-y)-tan(ix+y)\]
\[B= \frac{sin (ix+y-ix+y)}{2 cos (ix+y) cos (ix-y)}\]
\[=> B = \frac{sin 2y}{cos (2ix)+cos (2y)} = \frac{sin 2y}{cosh 2x + cos 2y}\]
is the required imaginary part of tanh z.
Thus we obtained real and imaginary parts of tanh(z) are sinh 2x/ (cosh (2x)+(cos(2y)) and sin 2y/(cosh 2x + cos 2y) respectively .