Question

tanh z is an analytic function? If analytic means how to prove this.

Answer 1

I assume you mean analytic function as a complex function (so a holomorphic function). We can rewrite tanh as follows:
\[ \tanh(a+bi)=\frac{e^{a+bi}-e^{-(a+bi)}}{e^{a+bi}+e^{-(a+bi)}} \]
Now either you can argue that since the complex exponential is analytic the expression to the right must be analytic and hence tanh is analytic. Or you can explicitly show that it is holomorphic by putting it into some form \(\tanh(a+bi)=u(a,b)+iv(x,y)\), i.e. split it into real and imaginary parts. Now to show that it is holomorphic, you can show that it satisfies the Cauchy-Riemann equations:
\[ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \]
and
\[ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.\]

Answer 2

One easiest way of breaking up \(\tanh(a+bi)\) to real and imaginary parts would probably be to use that it's sinh/cosh, then use the identities
\[ \sinh(a+bi)=\sinh(a)\cos(b)+i\cosh(a)\sin(b) \\
\cosh(a+bi)=\cosh(a)\cos(b)+i\sinh(a)\sin(b)\]
You would have to extend the resulting fraction by the complex conjugate of the denominator and then simplify, that should break it up into real and imaginary parts. Now show that the Cauchy-Riemann equations hold for the real and imaginary parts (\(u\) and \(v\)).

Answer 3

tanh x+i tanh y is an analytic or not. if not means where the point are not analytic becomes?