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MCQMediumJEE 2023Complex Numbers Basics

JEE Mathematics 2023 Question with Solution

Let the complex number z=x+iyz = x + iy be such that 2z−3i2z+i\frac{2z - 3i}{2z + i} is purely imaginary. If x+y2=0x + y^2 = 0, then y4+y2−yy^4 + y^2 - y is equal to:

  • A

    32\frac{3}{2}

  • B

    23\frac{2}{3}

  • C

    43\frac{4}{3}

  • D

    34\frac{3}{4}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: z=x+iyz = x + iy and 2z−3i2z+i\frac{2z - 3i}{2z + i} is purely imaginary. Also, x+y2=0x + y^2 = 0.

Find: The value of y4+y2−yy^4 + y^2 - y.

Since the given expression is purely imaginary, its real part must be zero:

Re⁡(2z−3i2z+i)=0\operatorname{Re}\left(\frac{2z - 3i}{2z + i}\right) = 0

Substitute z=x+iyz = x + iy:

2z−3i2z+i=2(x+iy)−3i2(x+iy)+i=2x+i(2y−3)2x+i(2y+1)\frac{2z - 3i}{2z + i} = \frac{2(x + iy) - 3i}{2(x + iy) + i} = \frac{2x + i(2y - 3)}{2x + i(2y + 1)}

Rationalizing the denominator,

2x+i(2y−3)2x+i(2y+1)=(2x+i(2y−3))(2x−i(2y+1))(2x+i(2y+1))(2x−i(2y+1))\frac{2x + i(2y - 3)}{2x + i(2y + 1)} = \frac{(2x + i(2y - 3))(2x - i(2y + 1))}{(2x + i(2y + 1))(2x - i(2y + 1))}

The denominator becomes

4x2+(2y+1)24x^2 + (2y + 1)^2

and the real part of the numerator is

4x2+(2y−3)(2y+1)4x^2 + (2y - 3)(2y + 1)

So,

4x2+(2y−3)(2y+1)=04x^2 + (2y - 3)(2y + 1) = 0

Expand and simplify:

4x2+4y2−4y−3=04x^2 + 4y^2 - 4y - 3 = 0

From x+y2=0x + y^2 = 0, we get

x=−y2x = -y^2

Substitute this into the equation:

4(−y2)2+4y2−4y−3=04(-y^2)^2 + 4y^2 - 4y - 3 = 0

which gives

4y4+4y2−4y−3=04y^4 + 4y^2 - 4y - 3 = 0

Now divide by 44:

y4+y2−y=34y^4 + y^2 - y = \frac{3}{4}

Therefore, the value of y4+y2−yy^4 + y^2 - y is 34\frac{3}{4}. The working leads to option C on the solution, but this value matches option D in the listed choices, so the correct option by value is D.

Answer Discrepancy Note

The solution derives

4y4+4y2−4y−3=04y^4 + 4y^2 - 4y - 3 = 0

which directly implies

y4+y2−y=34y^4 + y^2 - y = \frac{3}{4}

However, the same the solution incorrectly labels the correct option as C and also writes Correct Option: (3). Among the provided options, 34\frac{3}{4} is option D. Therefore, the mathematically defensible answer is D.

Common mistakes

  • Setting the whole fraction equal to zero instead of only its real part. A purely imaginary number need not be zero; only its real part must vanish. Use Re⁡(w)=0\operatorname{Re}(w)=0, not w=0w=0.

  • Using x=y2x = y^2 instead of x=−y2x = -y^2 from x+y2=0x + y^2 = 0. This sign error changes the final expression completely. Rearrange carefully before substitution.

  • Stopping at 4y4+4y2−4y−3=04y^4 + 4y^2 - 4y - 3 = 0 and not isolating the required quantity. The question asks for y4+y2−yy^4 + y^2 - y, so divide the equation by 44 and then solve for that expression.

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