this one is quiet subtle:

(i-i^(-1))^(-1)

=

(i+i^(-1))[(i-i^(-1))(i+i^(-1))]^(-1)

=(i+i^(-1))[i^2-1/i^2]^(-1)
=(i+i^(-1))[-1+1]^(-1)
=(i+i^(-1))(0)^(-1)
=undefined! (since 0^(-1) or in other words, 1/0 is undefined!)

(i-i^(-1))^(-1)
=i(i(i-i^(-1)))^(-1)
=i(i^2-1)^(-1)
=i(-2)^(-1)
=-i/2

so -i/2=undfined
so either -1/2 is undfined or i is undefined

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given an arbitary triangle ABC, draw angle bisector at A and perpendicular bisector and BC.

if the two lines are congruent, then you know that AB=AC.

case 2: let them meet at D. draw E,F,G such that DE, DF, DG perpendicular to BC, AC, AB respectively.

since BAD=CAD (angle bisector) and DGA=DFA=90, and AD=AD (common side) then we have triangle ADG is congruent to ADF. so AG=AF, GD=GF

also, we have BE=EC (bisector), and DEB=DEC=90 and ED=ED we have triangle BDE is congruent to triangle EDC

so BD=DC.

triangle BDG is also congruent to triangle CDF because GD=DF and BD=DC as shown above and BGD=CFD=90. so BG=FC

since AG=AF and GB=FC then AB=BC

do that to other two sides to get all triangles are equilateral