I am studying the principal square root function of complex numbers. On Wikipedia they present a complex number $z$ using polar coordinates as

\begin{equation} z = r \mathrm{e}^{i \varphi}, \quad r \ge 0, ~ -\pi < \varphi \le \pi. \end{equation}

Further, they define the principal square root of $z$ as

\begin{equation} \sqrt{z} = \sqrt{r} \mathrm{e}^{i \varphi/2}. \tag{1} \end{equation}

Continuing, it is mentioned that

The principal square root function is thus defined using the nonpositive real axis as a branch cut. The principal square root function is holomorphic everywhere except on the set of non-positive real numbers (on strictly negative reals it isn't even continuous).

I do not understand these two statements. My questions are

  1. Why is the principal square root function defined using the nonpositive real axis as a branch cut? It seems to me that for $z = \mathrm{e}^{i \pi}$, we obtain by equation $(1)$ the principal square root $\sqrt{z} = \sqrt{1} \mathrm{e}^{i \pi/2} = i$.
  2. Why is the principal square root function not continuous on the negative reals?
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    Consider $\sqrt{-1 + it},\; t \in \mathbb{R}$. Let $t$ approach $0$ through a) positive values, b) negative values. – Daniel Fischer Aug 5 '15 at 11:00
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    Your point 1 is correct, but the wikipedia article doesn't actually say that the function is not defined on the negative reals: it is and the formula you have given is correct. The statement about continuity is a bit misleading or even ambiguous: if you view it as a function $\mathbb{R}_{<0} \rightarrow \mathbb{C}$, it is continuous, but as Daniel Fischer points it is not continuous at negative reals when viewed as a function $\mathbb{C}\rightarrow\mathbb{C}$ – Rob Arthan Aug 5 '15 at 11:12
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    @DanielFischer I see that in case a) we obtain $\lim_{t \downarrow 0} \sqrt{-1 + it} = i$ and in case b) $\lim_{t \uparrow 0} \sqrt{-1 + it} = -i$. That explains why it is not continuous on the negative reals for the function $\sqrt{} : \mathbb{C} \to \mathbb{C}$. On the other hand, equation (1) can be viewed as a function of $\varphi$ in the sense $(-\pi,\pi] \to \mathbb{C}$. Is it not continuous when viewed like that? Or am I making a mistake in this reasoning? – Ritz Aug 5 '15 at 11:19
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    We have a topology on $\mathbb{C}$, and that is what is relevant for continuity. You have a bijection $(0,+\infty)\times (-\pi,\pi] \to \mathbb{C}\setminus \{0\}$ via $(r,\varphi) \mapsto r\cdot e^{i\varphi}$, and that bijection is continuous, but its inverse is not continuous. And since we're looking at functions whose domain is $\mathbb{C}$, even if we write complex numbers in polar form, it's the topology of $\mathbb{C}$ and not the topology of $(0,+\infty)\times (-\pi,\pi]$ that is relevant. – Daniel Fischer Aug 5 '15 at 11:33
up vote 0 down vote accepted

That is a convention that for principal value in general we must use $-\pi<\arg z<\pi$ for other values you can change the branch cut.

You will get different limits when approaching the negative x-axis from the upper half-plane and the lower half-plane, so by definition, the principal square root function is not continuous on the negative real.

The principle square root of a negative real number $-a$ is $\sqrt{a}i$. The principle square root of $a + bi$, where $b > 0$, is $x + yi$ where

$x = \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}}$

$y = \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}$

The principle square root of $a + bi$, where $b < 0$, is $x - yi$ where $x$ and $y$ have the same formulae as above.

The other square root is the negative of the principle square root.

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