Question

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

Answer 1

Unless I'm missing something:

`sqrt(z^2) = z " (primary root) " or +-z " (primary and secpndary roots)"`

Explanation:

I'm not sure why specifying `z in CC` is significant.

If `z = a+bi`
then
`color(white)("XXX")z^2 = a^2+2abi-b^2`

Any value (in `CC`), `hatz` for which
`color(white)("XXX") hatz^2 = a^2+2abi-b^2`
should be a square root of `z`

The two possible `hatz` values are
`color(white)("XXX")hatz = a+bi =z`
and
`color(white)("XXX")hatz = -a-bi = -z`

Answer 2

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) = sqrtabs(z)(cis(1/2 Arg(z)))`
where `Arg(z)` is the principal argument of `z`, which some take to be in `[0,2pi)` and others take to be in `(-pi, pi]`

So for radicand `z^2`, I would take

`sqrt(z^2) = sqrtabs(z^2)(cis(1/2 Arg(z^2)))`
where `Arg(z^2)` is the principal argument of `z^2`, which some take to be in `[0,2pi)` and others take to be in `(-pi, pi]`

Given a choice, I think I would prefer `Arg` in `(-pi, pi]`

Answer 3

If `Arg(z)` is defined to have range `(-pi, pi]`, then:

`sqrt(z^2) = { (z, Arg(z) in (-pi/2, pi/2]), (-z, Arg(z) in (-pi, -pi/2] uu (pi/2, pi]) :}`

Explanation:

If `Arg(z)` is defined to have range `[0, 2pi)`, then:

`sqrt(z^2) = { (z, Arg(z) in [0, pi)), (-z, Arg(z) in [pi, 2pi)) :}`

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 `Arg(z)` we choose determines where the discontinuity in `sqrt` occurs and the answer to whether `sqrt(-2i) = i - 1` or `1 - i`.