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hsbhatt

P(z) = z²+bz+c; b,c

C is such that |P(z)| = 1 whenever |z| = 1. Prove that b=c=0.

ayshwarya

a short cut 4 dis is 1st as it is mentioned ven |z| =1 lets take z=i

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

sprinkle

hsbhatt!

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!

hsbhatt

good soln by sprinkle. aishwarya pls note that b and c are complex

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