# complex analysis - differentiabiliity

The exact question is as follows:

Determine all points in the complex plane where the following function is differentiable.

$$f(z)=\frac{1}{z^2+iz+1}+\cosh(\sin(z))$$

I'm going to use the Cauchy Riemann equations to determine analyticity. The function is differentiable wherever it's analytic. To do so, I need to break up the function into its real and imaginary components, that is:

$$f(z)=f(x+iy)=u(x,y)+iv(x,y)$$

The first term in the summation is easy to deal with. However, I'm having trouble breaking down the cosh expression, hence this post.

$$\cosh(\sin(z))\:\: \ldots\ldots\ldots\:\:(1)$$ where $z \in\mathbb{C}$. My attempt is as follows:

Let $\sin(z)=\frac{i}{2}\left(e^{-iz}-e^{iz}\right)=-i\theta$. $$\cosh(\sin(z))=\cosh(-i\theta) \\ \cosh(-i\theta)=\cosh(i\theta)=\frac{1}{2}(e^{i\theta}+e^{-i\theta}),$$ where $\theta=\frac{1}{2}\left(e^{-iz}-e^{iz}\right).$

So after all that, I arrive at:

$$\cosh(\sin(z))=\cos(\theta) \\ where \:\: \theta \:\:is\:\: defined \:\:as\:\: above.$$

I think I have two options:

1. Show that the LHS expression of the sum is not differentiable when the denominator goes to zero. Argue that sin and cosh are both analytical over all C such that a composition of the two is also analytic.

2. Bite the bullet and use C-R to show analyticity. The RHS expression is very trickly.

• What is the question exactly? How do differentiate $\cosh \sin z$ (as a complex function)? – Travis Sep 7 '14 at 5:57
• What would happen if $z^2+iz+1=0?$ – graydad Sep 7 '14 at 7:11