What are real and imaginary parts of tanh z ?

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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 .


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