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I'm considering the following question:

Where is $f(z)=\sqrt{e^z+1}$, $z \in \mathbb{C}$, analytic? Find $f'(z)$ where it is analytic.

My approach has been to simply differentiate $f(z)$ to get $$f'(z)=\frac{\mathrm{e}^z}{2\sqrt{\mathrm{e}^{z}+1}}$$ and note that then $\mathrm{e}^z \neq -1$, giving $z \neq i(2k+1)\pi$, $k \in \mathbb{Z}$. But I realize that the square root isn't continuous on all of $\mathbb{C}$, so are there any intricacies that I should be paying attention to?

Looking forward to your replies!

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    $\begingroup$ Hint: $f(z)$ is the composition of three functions $f1(z)=e^z$, $f2(z)=z+1$, $f3(z)=\sqrt(z)$. if $g(z)$ and $h(h)$ are analytic in $z_0$ then $g(h(z)$ is. $\endgroup$ – miracle173 Sep 16 '13 at 7:32
  • $\begingroup$ We know that $mathrm{e}^z$ is entire, this $\mathrm{e}^z+1$ is entire. It can be proven that $\sqrt{z}$ is analytic on $\mathbb{C}$ by converting it to polar form and using the polar cauchy-riemann equations. Then it is analytic everywhere and we can differentiate and note that $\mathrm{e}^z \neq 1$ as written in my post above. Thanks! $\endgroup$ – Numbersandsoon Sep 16 '13 at 14:22
  • $\begingroup$ @ Bo Schmidt: You wrote"It can be proven that $\sqrt{z}$ is analytic on $\mathbb{C}$ by converting it to polar form and using the polar cauchy-riemann equations". Could you ground these words? $\endgroup$ – user64494 Sep 16 '13 at 15:25
  • $\begingroup$ We can write $z^{1/2}=\sqrt{r}\cos(\theta/2) + i\sqrt{r}\sin(\theta/2)$ for some complex number with modulus $r$ and argument $\theta$ (note that I'm unsure if the range of $\theta$ should be $[-\pi,\pi)$ or $[-2\pi,2\pi)$, it seems the first range only includes half of the complex plane which makes me reevaluate my previous statements). The polar C-R Eqs are $\partial u/\partial r=(1/r)\partial v / \partial \theta$ and $\partial v / \partial r = (-1/r) \partial u / \partial \theta$. Now, $\partial u/\partial r = (1/2\sqrt{r})\cos(\theta/2)$ etc. and the equations are seen to hold. $\endgroup$ – Numbersandsoon Sep 16 '13 at 22:51
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The term analytic often allows the function to be multivalued (i.e., to be a function of the path of analytic continuation). If this is the meaning of analytic in your post, then the answer is: "$f$ is analytic at every point except odd multiples of $\pi i$, and its derivative is given by the formula you found, where the branch of the square root is the same as the one used for $f$."

If you insist on a single-valued function (which I would call holomorphic, to disambiguate from the above), then one needs some branch cuts. There is no unique "canonical" choice of branch cuts. One just needs to cut every closed path for which the sum of winding numbers around odd multiples of $\pi$ is an odd number.

One way to do this is to remove line segments of length $2\pi$ connecting two consecutive multiples of the above form. That is, let $$\Omega = \mathbb C\setminus \bigcup_{n\in\mathbb Z} [(4n-1)\pi i, (4n+1)\pi i]$$ where the notation $[a,b]$ means the line segment from $a$ to $b$ in the complex plane. The cuts ensure that every closed curve in $\Omega$ that winds around one odd multiple of $\pi i$ also winds around another one, with the same index. As a result, $e^z+1$ travels an even number of time around the origin, which makes its square root single-valued.

Another option, perhaps a simpler one, is to cut the plane along horizontal half-lines starting at every odd multiple of $\pi i$. The remaining domain is simply-connected, so the monodromy theorem applies.

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