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

Question

Caleb Alexander · Accepted 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]) :}# 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)) :}# In general: 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#.

Theodore Amidon · Answer

I am not well versed in complex analysis.

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]#

Christian Antunez · Answer

Unless I'm missing something:

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

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#

Benjamin Achtenberg · Answer

undefined If $ z $ is a complex number, then $ \sqrt{z^2} $ can be simplified as follows:

$$ \sqrt{z^2} = \sqrt{(z)^2} = |z| $$

Where $ |z| $ denotes the magnitude or absolute value of the complex number $ z $. Therefore, if $ z $ is a complex number, $ \sqrt{z^2} $ simplifies to the absolute value of $ z $.