# Complex derivative of $\cos(x) \cosh(y)-i \sin(x) \sinh(y)$

The problem: Determine the derivative of the following function
$f(z)=\cos(x) \cosh(y)-i \sin(x) \sinh(y)$

The original exercise can be found at 2.17 (e) page 36

Should i try to rewrite the function in terms of $z=x+iy$ or is there some connection between the partial derivatives and the complex derivative that I am missing?

Edit: Thanks @Claude Leibovici
The equations needed are:
$(1) \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b)$
$(2) \cos(ic)=\cosh(c)$
$(3) \sin(id)=i\sinh(d)$

With (2) and (3) I can rewrite the cosh and sinh in terms of cos and sin.
$f(z)=\cos(x)\cos(iy)-\sin(x)\sin(iy)$

Then I can use (1) to combine it into one cos.
$f(z)=\cos(x+iy)=\cos(z)$

Therefore:
$f'(z)=-\sin(z)$

Observe that the form of your function recalls the addition formula for trigonometric functions; \begin{align*} \cos z &=\cos(x+iy)\\ &=\cos x\cos iy-\sin x\sin iy\\ &=\cos x\cosh y -i\sin x\sinh y\\ \end{align*} Thus the expression given is simply a reformulation of the complex cosine, i.e. $$f(z)=\cos z$$ thus the complex derivative of $f$ is $f'(z)=-\sin z$.

Hint

$$\cos(a+b)=\cos(a]\cos(b)-\sin(a)\sin(b)$$ $$\cos(i c)=\cosh(c)$$ $$\sin(id)=i\sinh(d)$$

All of that would make $f(z)$ very nice.

I am sure that you can take it from here.

Use Cauchy-Riemann (with the obvious $u,v,u_x,\dots$)

$$u_x = -\sin x \cosh y = v_y$$ $$u_y = \cos x \sinh y = - v_x$$

This shows that f is complex differentiable, and then you compute $$f'(z) = u_x+iv_x= \sin x \cosh y - i \cos x \sinh y.$$

• @surb: Thank you, fixed. – gammatester Feb 23 '16 at 11:28