|COMPLEXE NUMBERS| Show that?... thank you!

Question

Show that:
Is z ∈ ℤ, than is #z^-1 = z^- / |z|^2#
The z in the numerator is supposed to be the complex conjugate of z.
Thank you!

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Answer 1 · 2018-01-14T15:54:26

Yes

take #z= x + iy#
then #z^-1=1/(x + iy)#
#|z|^2=x^2 + y^2= (x+iy)(x-iy) = zz^_#

hence we get
RHS

#z^_ / (zz^_) = 1/z = z^-1# = LHS
hence proved
Hope u find it helpful :)