Checking where $\frac{1}{z^5-1}$ is differentiable

Question

I want to check where $\frac{1}{z^5-1}$ and $|z|\sin(z)$ are differentiable.

My question is:

is there any other (analytic) way to check it other than multiplying with its conjugate and +1 and doing a slow and painful process of expanding the binomial for the first one?

Also, any hints on the second one would be greatly appreciated!

Answer 1

For the first one, I would do as follows:

• since $z\mapsto z$ is differentiable and the product of differentiable functions is differentible, $z\mapsto z^5$ is differentiable;
• since the sum of differentiable functions is differentiable and since constant functions are differentiable, $z\mapsto z^5-1$ is differentiable;
• since the multiplicative inverse of a differentiable function is differentiable, $z\mapsto\frac1{z^5-1}$ is differentiable.

So, the answer is: the function is differentiable at every point of its domain, which is $\mathbb{C}\setminus\{\text{fifth roots of }1\}$.

Answer 2

For the first one, the answer by Santos is OK but it leaves out the five points at which the function blows up and is not differentiable, namely, at $$ \begin{array}{l} z = 1 \\ z = \frac14 (\sqrt{5}-1) + \sqrt{\frac18} i\sqrt{5+\sqrt{5}} \\ \frac14 (-\sqrt{5}-1) + \sqrt{\frac18} i\sqrt{5-\sqrt{5}} \\ -\frac14 (-\sqrt{5}-1) + \sqrt{\frac18} i\sqrt{5-\sqrt{5}} \\ \frac14 (\sqrt{5}-1) -\sqrt{\frac18} i\sqrt{5+\sqrt{5}} \end{array} $$ which are the five fifth roots of $1$.

The second function, $|z| \sin z$ is also subtle. Clearly it is differentiable at any $z \neq 0$ but does the function $\sin z$ go to zero rapidly enough to render it differentiable at $z = 0$? (For example, $|z|$ is not differentiable at $z=0$.)

I believe the answer is that $|z| \sin z$ is differentiable at $z=0$ for the same reason that the real function $x|x|$ has a well-defined derivative equal to $0$ at $x=0$. If that is right, then is the function $|z| \sin z$ analytic everywhere? That which certainly makes me uncomfortable, but there it is...

Answer 3

For the second one, it's probably easiest to use the Cauchy-Riemann conditions. If we write a complex-valued function as $f(x +iy) = u(x,y) + i v(x,y)$, then $f$ is complex-differentiable if and only if $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \& \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. $$ For this function, we have $$ f(x + iy) = |x + iy| \sin (x + iy) = \sqrt{x^2 + y^2} \left[ \sin x \cos (iy) + \cos x \sin(iy) \right] \\ = \sqrt{x^2 + y^2} \left[ \sin x \cosh y + i \cos x \sinh y \right]. $$ Take it from here. The conditions on $x$ and $y$ that emerge are not the easiest to solve, but assuming I did the problem correctly when I worked through it, there are a countably infinite number of points where the function is differentiable.