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Using triangle inequality,
|bz + c| <= |bz| + |c|
=> - |bz + c| >= - |bz| - |c| (multiply with -1 both sides)
=> |z2| - |bz + c| >= |z2| - |bz| - |c| ....... (i) (add |z2| both sides)
Again using triangle inequality,
|z2+bz+c| >= |z2| - |bz + c|
=> |z2+bz+c| >= |z2| - |bz| - |c| (using (i), if a>b, b>c then a>c)
when |z| = 1
1 >= 1 - |b| - |c| which is possible only when |b| = |c| = 0 => b = c = 0
Done!
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C is such that |P(z)| = 1 whenever |z| = 1. Prove that b=c=0.







hence p(z)=c-1+ib hence |p(z)| =1
so b^2 +c^2 -2c =0 hence b^2+c(c-2) =0
since b^2 is non -ve so b=0 c=0 c=2 r d so.lution der