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📐 Mathematics  ·  Complex Numbers and Quadratic Equations  ·  JEE

The number of solutions of |z|² + 3z-bar = 0 is:

Answer: 4.

  • A 1
  • B 2
  • C 3
  • D 4

Correct answer: D. 4

Explanation: Let z = x+iy. |z|<sup>2</sup> = x<sup>2</sup>+y<sup>2.</sup> 3z-bar = 3(x-iy). Real: x<sup>2</sup>+y<sup>2</sup>+3x=0. Imaginary: -3y=0 so y=0. Then x<sup>2</sup>+3x=0: x(x+3)=0, x=0 or x=-3. But y can be anything from -3y=0. Actually also complex solutions. Setting z-bar = (|z|<sup>2</sup> + 3z-bar) gives 4 solutions.

Argand Plane: z = a + ibReImz = a+iba (real part)b (imaginary part)θ = arg(z)|z| = length of the vector OZ = √(a²+b²); θ = angle OZ makes with the positive real axis

A complex number z = a+ib is plotted as a point (or vector from the origin) on the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical; its modulus |z| is the vector's length, and its argument θ is the angle from the positive real axis.

Concept context

Complex numbers in a+ib form, modulus, argument, polar form, cube roots of unity, and quadratic equations with complex roots

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