How to find cotangent of complex numbers

Question

I am having really small problem but don't know why Google search is not yielding fruitful results.

if

cotangent(x) = 1/tangent(x)

then how to find

cotangent(x+iy) 

Answer 1

It is given in Abramowitz/Stegun formula 4.3.58 or at http://dlmf.nist.gov/4.21.E40 $$\cot z = \cot(x + iy) = \frac{\sin 2x - i \sinh 2y}{\cosh 2 y - \cos 2x}$$

Answer 2

$$\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$$

Now $\cot(iy)=\dfrac{\cos(iy)}{\sin(iy)}=\dfrac{\cosh(y)}{i\sinh(y)}$

Answer 3

Hint: You can always express any trig function of a complex argument in terms of trig/hyperbolic functions of a real argument using the basic facts that $$\sin it =i\sinh t$$ $$\cos it =\cosh t$$ Note that these are completely analogous to the formulas $$\sin -t =-\sin t$$ $$\cos -t =\cos t$$

These, along with the normal sum formulas for sine and cosine, let you simplify any trig function of a complex number (because the other trig functions are expressable in terms of sine and cosine).