Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202429 Jan 2024Evening ShiftMathematicsComplex NumberActual

Let r and θ respectively be the modulus and amplitude of the complex number z = 2 - i 2 tan 5 π 8 , then ( r , θ ) is equal to

Options

  1. A2 sec 3 π 8 , 3 π 8
  2. B2 sec 3 π 8 , 5 π 8
  3. C2 sec 5 π 8 , 3 π 8
  4. D2 sec 11 π 8 , 11 π 8

Correct answer

A. 2 sec 3 π 8 , 3 π 8

Step-by-step solution

Given: z = 2 - i 2 tan 5 π 8 ⇒ z = 2 1 - i tan 5 π 8 ⇒ z = 2 1 - i sin 5 π 8 cos 5 π 8 ⇒ z = 2 cos 5 π 8 cos 5 π 8 - i sin 5 π 8 ⇒ z = 2 cos π - 3 π 8 cos π - 3 π 8 - i sin π - 3 π 8 ⇒ z = 2 - cos 3 π 8 - cos 3 π 8 - i sin 3 π 8 ⇒ z = 2 cos 3 π 8 cos 3 π 8 + i sin 3 π 8 ⇒ z = 2 sec 3 π 8 e i · 3 π 8 Now on comparing with z = z e i θ we get, ⇒ θ = 3 π 8 , r = 2 sec 3 π 8

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