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Algebra

Forum Expert
 Joined: 28 Feb 2007 Post: 2185
21 Jan 2008 14:47:15 IST
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Cmplx Numbers
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

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

Blazing goIITian

Joined: 5 Dec 2007
Posts: 704
21 Jan 2008 19:52:09 IST
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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

Hot goIITian

Joined: 10 Jan 2008
Posts: 107
21 Jan 2008 20:08:19 IST
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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!

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Joined: 28 Feb 2007
Posts: 2185
21 Jan 2008 21:23:26 IST
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good soln by sprinkle. aishwarya pls note that b and c are complex

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