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Algebra

Blazing goIITian

Joined:	13 Aug 2008
Post:	1313

8 Apr 2009 11:50:07 IST

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402

Complex Nos. simple one but im not getting :( :(

None

how do u do such Qsns ?

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Madmax

Blazing goIITian

Joined: 24 Dec 2007

Posts: 1111

8 Apr 2009 12:29:18 IST

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36 right??

I am representing conjugate of z by z' .
Second Condition: |z₁+2z₁+3z₃| = 6

Multiply z₁ by z₁'/z₁' , z₂ by z₂'/z₂' and z₃ by z₃'.
And use (z)(z') = |z|² for z₁ , z₂ and z₃ .

| |z₁|²/z₁' + 2|z₂|²/z₂' + 3|z₃|²/z₃' | = 6

| 1/z₁' + 8/z₂' + 27/z₃' | = 6

| (z₂'z₃' + 8z₃'z₁' + 27z₁'z₂')/(z₁'z₂'z₃') | = 6

Now take conjugate on both the sides :

| (z₂z₃ + 8z₃z₁ + 27z₁z₂)/(z₁z₂z₃) | = 6

| (z₂z₃ + 8z₃z₁ + 27z₁z₂) | / (|z₁|.|z₂|.|z₃|) = 6

| (z₂z₃ + 8z₃z₁ + 27z₁z₂) | = 6(|z₁|.|z₂|.|z₃|)

| (z₂z₃ + 8z₃z₁ + 27z₁z₂) | = 6(1)(2)(3) = 36

cheers!!

Here is a similar problem: http://www.goiit.com/posts/list/algebra-complex-nos1-11814.htm

Prakhar Banga

Blazing goIITian

Joined: 20 Dec 2008

Posts: 599

8 Apr 2009 12:49:26 IST

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Mod (z1 + 2z2 + 3z3) = 6.

Divide by Mod(z1z2z3) on the left and 6 on the right side ( As Mod(z1z2z3) = 6 )

Mod ( 1 / z2z3 + 2 / z1z3 + 3 / z1z2 ) = 1

Mod ( z2' z3' / (mod(z2))^2 (mod(z3))^2 + 2 z1' z3' / (mod(z1))^2 (mod(z3))^2+ 3 z2' z1' / (mod(z1))^2 (mod(z2))^2 ) = 1

Mod ( z2' z3' / 4*9 + 2 z1' z3' / 1*9 + 3 z2' z1' / 1*4 ) = 1

Multiply by 36

Mod ( z2' z3' + 8 z1' z3' + 27 z2' z1' ) = 36

So Mod ( z2 z3 + 8 z1 z3 + 27 z2 z1 ) = 36 ( As mod z' = mod z )

Option B.

Srujana

Blazing goIITian

Joined: 6 May 2007

Posts: 1025

8 Apr 2009 13:17:46 IST

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Here goes another method:

|x|=1 |y|=2 |z|=3

|x+2y+3z|=6

(x+2y+3z)(x'+2y'+3z')=36

2xy'+3xz'+2yx'+6yz'+3x'z+6yz'=-62

consider

|yz+8zx+27xy|

on squaring it

(yz+8zx+27xy)(y'z'+8z'x'+27x'y')

=36*(98+2xy'+3xz'+2yx'+6yz'+3x'z+6yz')

|Ans|^2=36*(36)

|Ans|=36

Mirka

Blazing goIITian

Joined: 13 Aug 2008

Posts: 1313

9 Apr 2009 08:30:33 IST

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Thank u verry much .....

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