Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Most Important Selected Qs for JEE AdvancedMathematicsComplex Number

Paragraph: For non-zero complex numbers z₁ and z₂ define the dot product and the cross product of z ₁ and z ₂ as follows Dot product z₁ z₂= |z₁ | |z₂ | =x₁ x₂+y₁ y₂= Re ( z ₁ z₂ ) Cross product z₁ z₂= |z₁ | |z₂ | =x₁ y₂-x₂ y₁=i m ( z ₁ z₂ ) Question: z₁ z₂==0 is equivalent to say

Options

  1. Az₂ z₁ is purely imaginary
  2. Bz₁ z₂ is purely real
  3. Cz₁=0 or z₂=0
  4. Dnone of these

Correct answer

C. z₁=0 or z₂=0

Step-by-step solution

aligned & z₁ z₂= |z₁ | |z₂ | =0 & Re ( z₁ z₂ )=0 & z ₁ z₂=i k & z₂ z₁ ( z z₁ )=i k & z₂ z₁ =i k aligned Also even if z₁=0 or z₂=0z₁ z₂=0

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 Most Important Selected Qs for JEE Advanced PYQs