33But let z=1/2+0i
|z|=1/2
|z-1|=1/2
|z+1|=3/2
so |z|=≠max(1/2,3/2)
if /z/ = max{ /z-1/ , /z+1/ } then, 1./z+conjugatez/=1\2
2.z+ conj. z =1
3./z+conj.z /=1
4.none
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8 Answers
62if /z/ = max{ /z-1/ , /z+1/ } then, '
think in terms of graph...
distance from zero is same as max of distance from 1 and -1
hence if Re(Z) > 0 then Re(Z) = 1/2
(This comes from the graph if you can figure out!!)
also if Re(Z) < 0 then Re(Z) = -1/2
3362oops galat jawaab...
the question should be min(|z-1|,|z+1)
probably..
33x2+y2=max((x-1)2+y2,(x+1)2+y2)
for x>0
max((x-1)2+y2,(x+1)2+y2) = (x+1)2+y2
so x2+y2=(x+1)2+y2
x=-1/2 but x>0 so no x
similarly for x<0 x=1/2 not possible..
so (d)
