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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.

I'm going to speak to my lecturer about this.

Thanks all.

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    $\begingroup$ What is the question exactly? How do differentiate $\cosh \sin z$ (as a complex function)? $\endgroup$ – Travis Sep 7 '14 at 5:57
  • $\begingroup$ What exactly are you attempting to do? There's no a priori necessity to simplify cosh(sin z). $\endgroup$ – anomaly Sep 7 '14 at 5:57
  • $\begingroup$ What would happen if $z^2+iz+1=0?$ $\endgroup$ – graydad Sep 7 '14 at 7:11
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    $\begingroup$ Sums, differences, products, quotients, compositions of differentiable functions are differentiable, so long as care is taken to stay within the domains of the various functions involved. The sine is defined and differentiable everywhere, so is the hyperbolic cosine, so they aren't going to cause any problems. $\endgroup$ – Gerry Myerson Sep 7 '14 at 7:40
  • $\begingroup$ Have you worked it out yet, Fractual? $\endgroup$ – Gerry Myerson Sep 9 '14 at 6:14

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