Given #z=((i)^i)^i# calculate #abs z# ?

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Answer 1 · 2017-04-29T04:41:13

#|z|=1#

#iota=sqrt(-1)# can also be written as #0+i# or

#cos(pi/2)+isin(pi/2)=e^(ipi/2)#

Hence #z=((iota)^iota)^iota#

= #((e^(ipi/2))^i)^i#

= #(e^(pi/2i^2))^i#

= #(e^(-pi/2))^i#

= #e^(-pi/2i)#

= #cos(-pi/2)+isin(-pi/2)#

= #0-i#

= #-i#

and #|z|=1#

Answer 2 · 2017-04-29T12:51:09

#1#

#absz =sqrt( bar z cdot z)#

now

#z=((i)^i)^i# and

#bar z = ((-i)^-i)^-i#

then

# bar z cdot z = ((-i)^-i)^-i((i)^i)^i = (-i)^-i(i)^i=(-i)i=1#

and

#absz = sqrt1 = 1#