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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Cmplx Numbers
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hsbhatt (3694)

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P(z) = z2+bz+c; b,c  C is such that |P(z)| = 1 whenever |z| = 1. Prove that b=c=0.
    
ayshwarya (241)

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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
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sprinkle (750)

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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 u
sing 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!  



Sorry for typing mistakes, please try to understand the symbols ...

-
Sprinkle
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hsbhatt (3694)

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

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