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How to compute the norm of a complex number under square root? Does the square of norm equal the norm of square:

$\|\sqrt z\|^2 = \|\sqrt {z^2}\|$?

Let $z = re^{i\theta}$, then $$\|\sqrt z\|^2 =\|\sqrt {re^{i\theta}}\|^2 = \|\sqrt r \sqrt {e^{i\theta}}\|^2 =\|\sqrt r {e^{1/2i\theta}}\|^2 = \|r {e^{i\theta}}\|.$$ And $$\|\sqrt {z^2}\|=\|\sqrt {(re^{i\theta})^2}\| = \|\sqrt {r^2e^{2i\theta}}\|= \|{re^{i\theta}}\|.$$

I hope this is correct? Thank you.

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  • $\begingroup$ $\sqrt z$ is ambiguous notation. $\endgroup$ – Pedro Tamaroff Jan 27 '14 at 10:27
  • $\begingroup$ er...? I meant the most common one, like $\sqrt{1 + 2i}$ @PedroTamaroff $\endgroup$ – 1LiterTears Jan 27 '14 at 10:28
  • $\begingroup$ Your calculations make no sense: the norm of a complex number is always a nonnegative real value. $\endgroup$ – heropup Jan 27 '14 at 10:29
  • $\begingroup$ Oh sorry @heropup, I meant to keep the norm. Thanks! $\endgroup$ – 1LiterTears Jan 27 '14 at 10:30
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    $\begingroup$ @1LiterTears There is no such thing as "the most common" squareroot of a complex number! $\endgroup$ – Pedro Tamaroff Jan 27 '14 at 10:40
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We claim that, for any $w \in \mathbb{R}$ and $z \in \mathbb{C}$, $$\|z^w\| = \|z\|^w.$$ Proof: if $\|z\| = 0$, then $z = 0$ and the desired condition is trivially satisfied. So suppose $\|z\| > 0$. Then $$\|z^w\| = \|r^w e^{iw\theta}\| = \|r^w\|\|e^{iw\theta}\| = \|r^w\| = \|r\|^w,$$ since $r > 0$. Also, $$\|z\|^w = \|r e^{i\theta}\|^w = \|r\|^w \|e^{i\theta}\|^w = \|r\|^w 1^w = \|r\|^w.$$ So they are equivalent, even if $z^w$ is multivalued.

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  • $\begingroup$ The definition of $z^\omega$ must be watched upon. This is where you lose generality. $\endgroup$ – AlexR Jan 27 '14 at 10:48
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    $\begingroup$ It doesn't matter: the magnitude of $z^w$ for any real $w$ is the same regardless of the branch chosen. $z^w = \exp(w \log z) = \exp(w \log |z| + i w \arg(z) + 2\pi i w k)$, and the magnitude is clearly $|z|^w$. $\endgroup$ – heropup Jan 27 '14 at 11:28
  • $\begingroup$ Still you can't define $z^\omega$ without choice of a branch. $\endgroup$ – AlexR Jan 27 '14 at 11:30
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    $\begingroup$ Of course you can. $z^w$ is a set that corresponds to a relation on $\mathbb C^2$. Nowhere is it required that such relations are single-valued, and certainly not for the purposes of this identity. You don't need to force a choice of branch, and especially not to uniquely define the norm, which as I have pointed out, is unique regardless of branch. $\endgroup$ – heropup Jan 27 '14 at 11:38
  • $\begingroup$ You should add that to your original answer (that $z^\omega$ need not necessarily be a holomorphic function on $\mathbb C$). $\endgroup$ – AlexR Jan 27 '14 at 11:47
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This is not correct. $|e^{i\theta}|=1 \quad \forall \theta\in\mathbb R$. Thus $$\Vert \sqrt z \Vert^2 = \Vert \sqrt r e^{i\frac\theta2} \Vert^2 = \Vert \sqrt r \Vert^2 = (\sqrt r)^2 = r$$ Where $r\ge 0$ by convention.
Note that $\sqrt z$ needs clarification. The usual definition excludes a line from $0$ to $\infty$, normally either $(-\infty, 0)$ or $(0, i\infty)$ then $$\sqrt z := \sqrt{|z|} e^{\frac12 i \arg z}$$


With regards to the edit:
Still you have the problem of defining the square root in the complex plane. If you chose a suitable definition, which limits you to a certain domain, equality will hold whenever $z$ and $z^2$ are in the domain of your square root.

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  • $\begingroup$ Hi Alex, thanks a lot. May I first ask - how can I deal with the simple case $\sqrt{1+i}$? $\endgroup$ – 1LiterTears Jan 27 '14 at 10:37
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    $\begingroup$ $(1+i)^2 = 2i$, so taking the square root to be $$\sqrt{\cdot} : \mathbb C \setminus (-\infty, 0) \to \mathbb C$$ Allows to write $|(1+i)^2| = 2 = {\sqrt 2}^2 = |1+i|^2$ $\endgroup$ – AlexR Jan 27 '14 at 10:40
  • $\begingroup$ Sorry for unclear - I am actually concerned about the case $\|\sqrt{1+i}\|$..? $\endgroup$ – 1LiterTears Jan 27 '14 at 10:43
  • $\begingroup$ In this case, the same definition of the square root will help you find $$\Vert \sqrt{1+i} \Vert = \sqrt[4]2$$ $\endgroup$ – AlexR Jan 27 '14 at 10:45
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    $\begingroup$ A small error: $\arg (1+i) = \frac\pi2 \neq \arg \frac\pi2$ $\endgroup$ – AlexR Jan 27 '14 at 10:49

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