 Branden Valentine

2022-02-01

If $z\in \mathbb{C}$ then what is $\sqrt{{z}^{2}}?$ liep3p

Step 1
I'm not sure why specifying $z\in \mathbb{C}$ is significant.
If $z=a+bi$
then
${z}^{2}={a}^{2}+2abi-{b}^{2}$
Any value (in for which
${\stackrel{^}{z}}^{2}={a}^{2}+2bi-{b}^{2}$
should be a square root of z
The two possible $\stackrel{^}{z}$ values are
$\stackrel{^}{z}=a+bi=z$
and
$\stackrel{^}{z}=-a-bi=-z$ Jordyn Horne

Step 1
I would take the principal square root of a complex number z, to be:
$\sqrt{z}=\sqrt{|z|}\left(cis\left(\frac{1}{2}arg\left(z\right\}\right)\right)$
where $arg\left(z\right)$ is the principal argument of z, which some take to be in and others take to be in
So for radicand ${z}^{2}$, I would take
$\sqrt{{z}^{2}}=\sqrt{|{z}^{2}|}\left(cis\left(\frac{1}{2}arg\left({z}^{2}\right)\right)\right)$
where $\mathrm{arg}\left({z}^{2}\right)$ is the principal argument of ${z}^{2}$, which some take to be in and others take to be in
Given a choice, would prefer $arg$ in $\left(-\pi ,\pi \right]$

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