Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202523 Jan 2025Evening ShiftMathematicsComplex NumberActual

Let , be the roots of the equation x^2-a x-b=0 with Im ( ) Im ( ) . Let P_n= ^n- ^n . If P ₃=-5 7 i, P ₄=-3 7 i, P ₅=11 7 i and P ₆=45 7 i , then | ^4+ ^4 | is equal to .

Correct answer

0

Step-by-step solution

aligned & + = a =- b & P ₆= aP ₅+ bP ₄ & 45 7 i = a 11 7 i + b (-3 7 ) i & 45=11 a -3 ~b aligned and aligned & P₅= aP ₄+b P₃ & 11 7 i = a (-3 7 i )+ b (-5 7 i ) & 11=-3 a -5 ~b & a =3, ~b =-4 & | ^4+ ^4 |= ( ^4- ^4 )^2+4 ^4 ^4 & = -63+4.4^4 & = -63+1024 = 961 =31 aligned

Practice Complex Number on Quantrex Academy →

More from Complex Number

The number of values of z C , satisfying the equations |z-(4+8i)|= 10 and |z-(3+5i)|+|z-(5+11i)|=4 5 , is: 2026Let S = z C : z^2 + 6 ,iz - 3 = 0 . Then _ z S z^8 is equal to : 2026Let the set of all values of k R such that the equation z( z + 2 + i) + k(2 + 3i) = 0 , z C , has at least one solution, be the interval [ , ] . Then 9( + ) is equal to: 2026Let z₁, z₂ C be the distinct solutions of the equation z^2 + 4z - (1 + 12i) = 0 . Then |z₁|^2 + |z₂|^2 is equal to : 2026Let S= z C : z^2+4z+16=0 . Then _ z S |z+ 3 i|^2 is equal to: 2026Let z be a complex number such that |z+2| = |z-2| and ( z+3 z-i ) = 4 . Then |z|^2 is equal to: 2026Let the circles C₁ : |z| = r and C₂ : |z - 3 - 4i| = 5 , z C , be such that C₂ lies within C₁ . If z₁ moves on C₁ , z₂ moves on C₂ and |z₁ - z₂| = 2 , then |z₁ - z₂| is equal to: 2026Let x and y be real numbers such that 50 ( 2x 1+3i - y 1-2i ) = 31 + 17i , i = -1 . Then the value of 10(x - 3y) is : 2026 Full Complex Number list All JEE Main PYQs