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JEE Main 2019
Complex Numbers
Complex Numbers
Easy

Question

If z=x+iy,xy0z=x+i y, x y \neq 0, satisfies the equation z2+izˉ=0z^2+i \bar{z}=0, then z2\left|z^2\right| is equal to :

Options

Solution

Key Concepts and Formulas

  • Modulus of a complex number: If z=x+iyz = x + iy, then z=x2+y2|z| = \sqrt{x^2 + y^2}.
  • Properties of the modulus:
    • z1z2=z1z2|z_1 z_2| = |z_1| |z_2|
    • zn=zn|z^n| = |z|^n
    • zˉ=z|\bar{z}| = |z|
    • i=1|i| = 1
  • Given xy0xy \neq 0, then x0x \neq 0 and y0y \neq 0, which implies z0z \neq 0 and thus z0|z| \neq 0.

Step-by-Step Solution

1. Rearrange the given equation. We are given z2+izˉ=0z^2 + i\bar{z} = 0. To prepare for applying the modulus, isolate the terms: z2=izˉz^2 = -i\bar{z} Explanation: Isolating the terms involving zz on one side allows us to apply the modulus operation to both sides and simplify the equation.

2. Apply the modulus to both sides of the equation. Take the modulus of both sides: z2=izˉ|z^2| = |-i\bar{z}| Explanation: Applying the modulus to both sides allows us to work with magnitudes, transforming the complex equation into a real-valued equation.

3. Simplify the moduli using relevant properties.

  • Left Hand Side (LHS): Using the property zn=zn|z^n| = |z|^n: z2=z2|z^2| = |z|^2 Explanation: This simplifies the LHS to the square of the modulus of zz.

  • Right Hand Side (RHS): Using the property z1z2=z1z2|z_1 z_2| = |z_1| |z_2|: izˉ=izˉ|-i\bar{z}| = |-i| |\bar{z}| Using i=i=1|-i| = |i| = 1: izˉ=1zˉ|-i| |\bar{z}| = 1 \cdot |\bar{z}| Using zˉ=z|\bar{z}| = |z|: 1zˉ=z1 \cdot |\bar{z}| = |z| Therefore, the RHS simplifies to z|z|. Explanation: We systematically applied the modulus properties. The modulus of a product was broken down, the modulus of i-i was found to be 1, and the modulus of the conjugate was replaced with the modulus of zz, simplifying the entire RHS to z|z|.

4. Equate the simplified expressions and solve for z|z|. From Step 3, we have z2=z|z|^2 = |z|. Rearrange the equation: z2z=0|z|^2 - |z| = 0 Factor out z|z|: z(z1)=0|z|(|z| - 1) = 0 This gives us two possible solutions for z|z|: z=0orz1=0    z=1|z| = 0 \quad \text{or} \quad |z| - 1 = 0 \implies |z| = 1 Explanation: We transformed the equation into a solvable form involving z|z|, leading to two potential values.

5. Validate the solutions for z|z| using the given conditions. The problem states that z=x+iyz = x + iy with xy0xy \neq 0. If xy0xy \neq 0, then x0x \neq 0 and y0y \neq 0. If z=0|z| = 0, then x2+y2=0\sqrt{x^2 + y^2} = 0, which implies x2+y2=0x^2 + y^2 = 0. This can only be true if x=0x = 0 and y=0y = 0. However, x=0x = 0 and y=0y = 0 contradicts the given condition xy0xy \neq 0. Therefore, z=0|z| = 0 is not a valid solution. The only valid solution is z=1|z| = 1. Explanation: The condition xy0xy \neq 0 means zz cannot be the origin (0+0i0+0i). Since z|z| represents the distance from the origin, z=0|z|=0 would mean z=0z=0, which is disallowed.

6. Calculate the required value z2|z^2|. The question asks for the value of z2|z^2|. Using the property z2=z2|z^2| = |z|^2, and since we found z=1|z| = 1: z2=(1)2=1|z^2| = (1)^2 = 1 Thus, z2=1|z^2| = 1.

Common Mistakes & Tips

  • Always check given conditions: The condition xy0xy \neq 0 was crucial here to eliminate the extraneous solution z=0|z|=0.
  • Modulus properties: Memorize and understand how to apply the modulus properties to simplify calculations.
  • Distinguish between zz and z|z|: zz is a complex number, while z|z| is its magnitude (a non-negative real number).

Summary

This problem demonstrates how to solve complex number equations by applying the modulus operation and leveraging the properties of complex moduli. We transformed the complex equation into a real-valued equation, solved for z|z|, and then used the given condition xy0xy \neq 0 to discard the extraneous solution. The final answer is z2=1|z^2| = 1.

Final Answer

The final answer is \boxed{1}, which corresponds to option (D).

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