What is the conjugate of #i^2#?

1 Answer
Nov 25, 2017

Conjugate of #i^2# is #i^2# itself and not #-i^2#

Explanation:

Conjugate of a complex number #a+ib# is #a-ib#

as #i^2=-1+i0#,

its conjugate is #-1-i0# or #-1# i.e. #i^2# - in case you wish to write it this way.

Hence conjugate of #i^2# is #i^2# itself and not #-i^2#

Also if you mark a complex number #z# in Argand plane and its conjugate, the two are reflection of each other on real number line.

As #i^2=-1#, it lies on real number line and it will be its own reflection.