# Define square root function over the complex numbers.

Define a holomorphic function $f\colon\mathbb{C}\setminus[-1,1] \longrightarrow \mathbb{C}$ such that $\forall z \in \mathbb{C}\setminus[-1,1] \ \left( (f(z))^{2} = z^{2} - 1\right)$ and $f(2)=\sqrt{3}$.

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just try square root of what is given to you, i.e. $e^{\frac{1}{2}\log(z^2-1)}$ – Golob Nov 14 '12 at 9:34
"The"? There are two square roots. Not only that, where sould the logarithm be defined? – Gong Nov 14 '12 at 9:48

I would try the following. Define:

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

where $x,y,u(x,y)$ and $v(x,y)$ are real. Now impose Cauchy-Riemman equations

$$u_{x}(x,y)=v_{y}(x,y)\\ u_{y}(x,y)=-v_{x}(x,y)$$

As there are differential equations you might expect to have one undetermined constant. It will be fixed by $f(2)=\sqrt{3}$

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I can't see how to make this work. – Gong Nov 14 '12 at 13:30

Let $\operatorname{Log} z$ denote the principal branch of the complex logarithm, i.e. the holomorphic choice of logarithm that is defined everywhere except on the negative real axis.

Note that $z^2-1 = z^2(1-\frac{1}{z^2})$ and the second factor is negative real if and only if $z \in [-1,1]$. This observation indicates that our "square root" should be

$$g(z) = z\sqrt{1-\frac{1}{z^2}} = z e^{\frac12 \operatorname{Log}(1-\frac{1}{z^2})}.$$

Make sure to check the details for yourself.

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"z2−1=z2(1−1z2) and the second factor is negative real if and only if z∉[−1,1]", this isn't true. For instance letting z=0.5 contradicts one of the directions, while z=i contradcits the other. – Gong Nov 14 '12 at 13:29
Sorry, it was supposed to the the other way around of course. – mrf Nov 14 '12 at 13:47
Sorry for not realizing that, it was obvious. Thank you very much. – Gong Nov 14 '12 at 17:03

Consider the square root function defined on $\mathbb C \setminus \mathbb R_-$ such that $\sqrt{1}=1$ (so it's an extension of the usual square root defined on $\mathbb R_+^*$). Define $f_1(z) = \sqrt{z-1} \sqrt{z+1}$ on $\mathbb C \setminus (- \infty ; 1]$ and $f_2(z) = f_1(-z)$ on $\mathbb C \setminus [- 1 ; \infty)$.

For $z \in \mathbb C \setminus \mathbb R$, we have $f_1(z)^2 = z^2-1 = (-z)^2-1 = f_2(z)^2$ so we must have $f_1/f_2 = \pm 1$ on each half-plane. Since $f_2(z)/f_1(z) = f_1(-z)/f_2(-z)$, we must have that $f_1/f_2$ is constant on $\mathbb C \setminus \mathbb R$. It turns out that $f_1 = - f_2$, so you can glue $f_1$ and $-f_2$ together to obtain a holomorphic function $f$ on $\mathbb C \setminus [-1 ; 1]$ satisfying $f(z)^2 = z^2-1$

Then you check that $f(2) = f_1(2) = \sqrt 1 \sqrt 3 = \sqrt 3$

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"Consider the square root function defined on C∖R− such that 1√=1", I got problems with that right there. What function is that? – Gong Nov 14 '12 at 13:31
it is the only continuous function $\sqrt{.}$ such that $(\sqrt{z})^2 = z$ and $\sqrt{1}= +1$. You can also define it by choosing $f(re^{it}) = \sqrt r e^{it/2}$ whenever $- \pi < t < \pi$. Or you can define it with a differential equation : it is the only holomorphic function $g$ such that $2g(z)g'(z) = 1$ and $g(1)=1$. – mercio Nov 14 '12 at 14:12

For any path between $2$ and $z$ that doesn't intersect the real line between $+1$ and $-1$, define $$f(z)=\frac{\log(3)}{2}+\int_2^z\frac{\zeta\,\mathrm{d}\zeta}{\zeta^2-1}$$ Since $f'(z)=\frac{z}{z^2-1}=\frac{1/2}{z-1}+\frac{1/2}{z+1}$, after accounting for the constant of integration at $z=2$, we get that $f(z)=\frac12\log(z^2-1)$ .

Since the sum of the residues of $\frac{\zeta}{\zeta^2-1}=\frac{1/2}{\zeta-1}+\frac{1/2}{\zeta+1}$ at $\zeta=+1$ and $\zeta=-1$ is $1$, the difference of the integral between two different paths that don't intersect the real line between $+1$ and $-1$ must be an integral multiple of $2\pi i$. Thus, $e^f$ is the same over both paths.

Therefore, $e^{f(z)}=\sqrt{z^2-1}$ is well-defined independent of the path taken.

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Wow. How do you even get such ideas? Amazing. Thanks. – Gong Nov 14 '12 at 20:51
It is a standard use of contour integration; define $\log$ using $\int\frac{\mathrm{d}z}{z}$ and use the right branch cuts to make the function well-defined. – robjohn Nov 14 '12 at 23:05