Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202431 Jan 2024Evening ShiftMathematicsComplex NumberActual

Let z 1 and z 2 be two complex number such that z 1 + z 2 = 5 and z 1 3 + z 2 3 = 20 + 15 i . Then z 1 4 + z 2 4 equals-

Options

  1. A30 3
  2. B75
  3. C15 15
  4. D25 3

Correct answer

B. 75

Step-by-step solution

Given: z 1 + z 2 = 5 & z 1 3 + z 2 3 = 20 + 15 i Now, solving z 1 3 + z 2 3 = 20 + 15 i ⇒ z 1 + z 2 3 - 3 z 1 z 2 z 1 + z 2 = 20 + 15 i ⇒ 5 3 - 3 z 1 z 2 5 = 20 + 15 i ⇒ 125 - 15 z 1 z 2 = 20 + 15 i ⇒ z 1 z 2 = 7 - i Now, finding z 1 4 + z 2 4 = z 1 2 + z 2 2 2 - 2 z 1 z 2 2 ⇒ z 1 4 + z 2 4 = z 1 + z 2 2 - 2 z 1 z 2 2 - 2 z 1 z 2 2 ⇒ z 1 4 + z 2 4 = 25 - 2 7 - i 2 - 2 7 - i 2 ⇒ z 1 4 + z 2 4 = 11 + 2 i 2 - 2 48 - 14 i ⇒ z 1 4 + z 2 4 = 121 - 4 + 44 i - 96 + 28 i ⇒ z 1 4 + z 2 4 = 21 + 72 i ⇒ z 1 4 + z 2 4 = 21 2 +

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