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JEE Main 2024
Complex Numbers
Complex Numbers
Hard

Question

Let the complex number z=x+iyz = x + iy be such that 2z3i2z+i{{2z - 3i} \over {2z + i}} is purely imaginary. If x+y2=0{x} + {y^2} = 0, then y4+y2y{y^4} + {y^2} - y is equal to :

Options

Solution

Key Concepts and Formulas

  • Purely Imaginary Numbers: A complex number ww is purely imaginary if w=wˉw = -\bar{w}, which is equivalent to w+wˉ=0w + \bar{w} = 0.
  • Complex Conjugate Properties:
    • z1±z2=z1ˉ±z2ˉ\overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2}
    • z1/z2=z1ˉ/z2ˉ\overline{z_1 / z_2} = \bar{z_1} / \bar{z_2}
    • ki=ki\overline{ki} = -ki for a real constant kk.
  • Modulus of a Complex Number: If z=x+iyz = x + iy, then z2=zzˉ=x2+y2|z|^2 = z\bar{z} = x^2 + y^2.

Step-by-Step Solution

1. Define the complex expression and apply the purely imaginary condition. Let w=2z3i2z+iw = \frac{2z - 3i}{2z + i}. We are given that ww is purely imaginary. Why: By definition, a complex number ww is purely imaginary if and only if w=wˉw = -\bar{w}, which can be rewritten as w+wˉ=0w + \bar{w} = 0. This allows us to convert the problem into an algebraic equation involving zz and its conjugate zˉ\bar{z}. 2z3i2z+i+(2z3i2z+i)=0\frac{2z - 3i}{2z + i} + \overline{\left(\frac{2z - 3i}{2z + i}\right)} = 0

2. Apply properties of complex conjugates. Using the properties of complex conjugates: (2z3i2z+i)=2z3i2z+i=2zˉ3i2zˉ+i=2zˉ+3i2zˉi\overline{\left(\frac{2z - 3i}{2z + i}\right)} = \frac{\overline{2z - 3i}}{\overline{2z + i}} = \frac{2\bar{z} - \overline{3i}}{2\bar{z} + \overline{i}} = \frac{2\bar{z} + 3i}{2\bar{z} - i} Why: Taking the conjugate of the entire fraction simplifies the expression by applying the conjugate operation to the numerator and denominator separately. It's crucial to correctly find the conjugate of terms like 3i-3i (which is +3i+3i) and +i+i (which is i-i). Substituting this back into our condition: 2z3i2z+i+2zˉ+3i2zˉi=0\frac{2z - 3i}{2z + i} + \frac{2\bar{z} + 3i}{2\bar{z} - i} = 0

3. Clear the denominators and expand the terms. Multiply both sides by (2z+i)(2zˉi)(2z + i)(2\bar{z} - i) to eliminate the fractions. Why: This is a standard algebraic step to simplify equations with fractions, allowing us to work with polynomials. (2z3i)(2zˉi)+(2zˉ+3i)(2z+i)=0(2z - 3i)(2\bar{z} - i) + (2\bar{z} + 3i)(2z + i) = 0 Now, expand both products:

  • First product: (2z3i)(2zˉi)=4zzˉ2zi6izˉ+3i2=4z22iz6izˉ3(2z - 3i)(2\bar{z} - i) = 4z\bar{z} - 2zi - 6i\bar{z} + 3i^2 = 4|z|^2 - 2iz - 6i\bar{z} - 3
  • Second product: (2zˉ+3i)(2z+i)=4zˉz+2izˉ+6iz+3i2=4z2+2izˉ+6iz3(2\bar{z} + 3i)(2z + i) = 4\bar{z}z + 2i\bar{z} + 6iz + 3i^2 = 4|z|^2 + 2i\bar{z} + 6iz - 3 Why: Careful expansion of products is essential to avoid algebraic errors. Recognizing zzˉ=z2z\bar{z} = |z|^2 simplifies the expression by converting complex number products into real magnitudes.

4. Combine and simplify the expanded terms. Add the results from the two products: (4z22iz6izˉ3)+(4z2+2izˉ+6iz3)=0(4|z|^2 - 2iz - 6i\bar{z} - 3) + (4|z|^2 + 2i\bar{z} + 6iz - 3) = 0 Combine like terms: 8z2+4iz4izˉ6=08|z|^2 + 4iz - 4i\bar{z} - 6 = 0 Why: Grouping and combining like terms makes the expression more manageable and moves us closer to an equation in terms of xx and yy.

5. Substitute z=x+iyz = x + iy and z2=x2+y2|z|^2 = x^2 + y^2. We are given z=x+iyz = x + iy. Therefore, zˉ=xiy\bar{z} = x - iy and z2=x2+y2|z|^2 = x^2 + y^2. Substitute these into the simplified equation: 8(x2+y2)+4i(x+iy)4i(xiy)6=08(x^2 + y^2) + 4i(x + iy) - 4i(x - iy) - 6 = 0 Expand the terms: 8(x2+y2)+4ix4y4ix4y6=08(x^2 + y^2) + 4ix - 4y - 4ix - 4y - 6 = 0 The terms 4ix4ix and 4ix-4ix cancel out: 8(x2+y2)8y6=08(x^2 + y^2) - 8y - 6 = 0 Divide the entire equation by 2 to simplify: 4(x2+y2)4y3=04(x^2 + y^2) - 4y - 3 = 0 Why: By substituting z=x+iyz=x+iy and zˉ=xiy\bar{z}=x-iy, we convert the complex number equation into an equation solely involving real variables xx and yy, which is easier to work with. This also utilizes the definition of the modulus squared, z2=x2+y2|z|^2 = x^2+y^2.

6. Utilize the second given condition. We are given the condition x+y2=0x + y^2 = 0, so x=y2x = -y^2. Why: This condition provides a link between xx and yy, allowing us to reduce the number of variables in our equation and solve for a specific expression involving only yy. Substitute x=y2x = -y^2 into the equation 4(x2+y2)4y3=04(x^2 + y^2) - 4y - 3 = 0: 4((y2)2+y2)4y3=04((-y^2)^2 + y^2) - 4y - 3 = 0 4(y4+y2)4y3=04(y^4 + y^2) - 4y - 3 = 0 4y4+4y24y3=04y^4 + 4y^2 - 4y - 3 = 0

7. Isolate the required expression. We need to find the value of y4+y2yy^4 + y^2 - y. Notice that 4y4+4y24y4y^4 + 4y^2 - 4y is 44 times the expression we are looking for. Rearrange the equation: 4(y4+y2y)=34(y^4 + y^2 - y) = 3 Divide by 4: y4+y2y=34y^4 + y^2 - y = \frac{3}{4} Why: By carefully observing the algebraic form of the equation and the desired expression, we can directly solve for it without needing to find the specific value of yy. This often simplifies calculations and avoids dealing with roots.

Common Mistakes & Tips:

  • Conjugate Errors: Mistakes in taking the conjugate, especially with terms involving 'i', are common. Remember i=i\overline{i} = -i.
  • Algebraic Simplification: Be careful with signs when expanding and simplifying.
  • Using w+wˉ=0w + \bar{w} = 0: This is the key to solving purely imaginary problems.

Summary This problem uses the properties of complex conjugates and the definition of purely imaginary numbers to find the value of an expression. Substituting x=y2x=-y^2 allowed us to directly compute the target expression y4+y2yy^4 + y^2 - y. The final answer is 34\frac{3}{4}.

Final Answer The final answer is 34\boxed{\frac{3}{4}}, which corresponds to option (C).

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