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

3 Answers
Sep 26, 2015

Answer:

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#

Sep 26, 2015

Answer:

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]#

Sep 26, 2015

Answer:

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#.