Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020MathematicsComplex NumberPractice

The real part of the complex number z satisfying z - 1 - 2 i ≤ 1 and having the least positive argument, is

Options

  1. A4 5
  2. B8 5
  3. C6 5
  4. D7 5

Correct answer

B. 8 5

Step-by-step solution

Here, z - 1 - 2 i = 1 represents a circle with centre 1 , 2 and radius 1 unit. The complex number z = x + i y satisfying the given inequality and having the least positive argument is the point of contact of the tangent from the origin to the circle with the least positive slope. From the diagram, tan ⁡ ϕ = 1 2 ∴ tan ⁡ 2 ϕ = 2 tan ϕ 1 - tan 2 ϕ = 4 3 For the least positive argument, arg z = θ (let) tan ⁡ θ = t a n π 2 - 2 ϕ = cot 2 ϕ = 3 4 Also, f

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 NTA Abhyas JEE Main PYQs