IIT JEE , AIEEE Forum - Mathematics , Algebra

pannu89

Dear Sir

Is there a technique to solve problems based on roots of complex equations?

For example

Q. If

₁,

₂,

₃,

₄ be 4 real numbers then any root of the equation

sin

₁z³+sin

₂z²+ sin

₃z +sin

₄ = 3

lying in the unit circle |z| = 1, satisfies the inequality

(a) |z| < 2/3 (b) |z| > 2/3 (c) |z| < 1/2 (d) |z| >1/2

I would be obliged if anyone answers my question

Pranav Kumar

hit_ur_heart

there isnt much to do in this , proceed this way..................

sin(x1)z^3 + sin(x2)z^2 + sin(x3)z + sin(x4) = 3
taking modulus on both side, we get
|sin(x1)z^3 + sin(x2)z^2 + sin(x3)z + sin(x4)| = |3|
but
|sin(x1)z^3 + sin(x2)z^2 + sin(x3)z + sin(x4)| <= |sin(x1)z^3| + |sin(x2)z^2| + |sin(x3)z| + |sin(x4)| .

but as we know that |sin(theta)| <= 1, for any value of theta.
=> 3 <= |z^3| + |z^2| +|z| + 1
=> |z|^3 + |z|^2 + |z| - 2 >= 0
now |z| is non-negative real number ,

Consider the function f(x) = x^3 + x^2 + x - 2
as this function will be strictly increasing ( verify it) , there will be only one root , let it be "r".

=> (|z| -r) ( some quadratic function of |z| with no real root ) >= 0
=> |z| >= r
Since maix value of |z| given in option is 2/3 , but f( 2/3) < 0
=> r > 2/3
=> |z| > 2/3
If u have any query regarding any step , u can post back

farazkhan

Good complete solution.

puneet

hiiii

hit_ur heart ... it is a very good solution .... completely correct

take a salute from my side

cheers

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