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sriram.a

A:If lzl=Max{lz-1l,lz+1l},thenlz+ $\bar{z}$ l=1

R:Max{a,b}

me

here z is given by --
lzl=lz-1l when lz-1l > lz+1l
this represent line x=1/2
hence z=1/2 + pi , for real p
thus z+conjugate = 1

and when lz+1l > lz-1l , lzl=lz+1l
this now represent line x=-1/2
thus z+conjugate = -1

thus in both cases lz+conjugatel=1
so A is correct.

R is also correct but not enough to state A.

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A:If lzl=Max{lz-1l,lz+1l},thenlz+l=1

R:Max{a,b}

A:If lzl=Max{lz-1l,lz+1l},thenlz+ $\bar{z}$ l=1