# If z in CC then what is sqrt(z^2)?

##### 3 Answers
Sep 26, 2015

Unless I'm missing something:

$\sqrt{{z}^{2}} = z \text{ (primary root) " or +-z " (primary and secpndary roots)}$

#### Explanation:

I'm not sure why specifying $z \in \mathbb{C}$ is significant.

If $z = a + b i$
then
$\textcolor{w h i t e}{\text{XXX}} {z}^{2} = {a}^{2} + 2 a b i - {b}^{2}$

Any value (in $\mathbb{C}$), $\hat{z}$ for which
$\textcolor{w h i t e}{\text{XXX}} {\hat{z}}^{2} = {a}^{2} + 2 a b i - {b}^{2}$
should be a square root of $z$

The two possible $\hat{z}$ values are
$\textcolor{w h i t e}{\text{XXX}} \hat{z} = a + b i = z$
and
$\textcolor{w h i t e}{\text{XXX}} \hat{z} = - a - b i = - z$

Sep 26, 2015

I am not well versed in complex analysis.

#### Explanation:

I would take the principal square root of a complex number $z$, to be:

$\sqrt{z} = \sqrt{\left\mid z \right\mid} \left(c i s \left(\frac{1}{2} A r g \left(z\right)\right)\right)$
where $A r g \left(z\right)$ is the principal argument of $z$, which some take to be in $\left[0 , 2 \pi\right)$ and others take to be in $\left(- \pi , \pi\right]$

So for radicand ${z}^{2}$, I would take

$\sqrt{{z}^{2}} = \sqrt{\left\mid {z}^{2} \right\mid} \left(c i s \left(\frac{1}{2} A r g \left({z}^{2}\right)\right)\right)$
where $A r g \left({z}^{2}\right)$ is the principal argument of ${z}^{2}$, which some take to be in $\left[0 , 2 \pi\right)$ and others take to be in $\left(- \pi , \pi\right]$

Given a choice, I think I would prefer $A r g$ in $\left(- \pi , \pi\right]$

Sep 26, 2015

If $A r g \left(z\right)$ is defined to have range $\left(- \pi , \pi\right]$, then:

$\sqrt{{z}^{2}} = \left\{\left(z , A r g \left(z\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right]\right) , \left(- z , A r g \left(z\right) \in \left(- \pi , - \frac{\pi}{2}\right] \cup \left(\frac{\pi}{2} , \pi\right]\right)\right.$

#### Explanation:

If $A r g \left(z\right)$ is defined to have range $\left[0 , 2 \pi\right)$, then:

$\sqrt{{z}^{2}} = \left\{\left(z , A r g \left(z\right) \in \left[0 , \pi\right)\right) , \left(- z , A r g \left(z\right) \in \left[\pi , 2 \pi\right)\right)\right.$

Roughly speaking:

If $z$ in Q1, then $\sqrt{{z}^{2}} = z$

If $z$ in Q3, then $\sqrt{{z}^{2}} = - z$

If $z$ is in Q2 or Q4 then it is not obvious whether $\sqrt{{z}^{2}}$ is $z$ or $- z$.

Which definition of $A r g \left(z\right)$ we choose determines where the discontinuity in sqrt occurs and the answer to whether $\sqrt{- 2 i} = i - 1$ or $1 - i$.