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NVAMediumJEE 2023Complex Numbers Basics

JEE Mathematics 2023 Question with Solution

Let S={z∈C−{i,2i}:z2+8iz−15z2−3iz−2∈R}S = \left\{z \in \mathbb{C} - \{i, 2i\}: \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \right\}. If α−1311i∈S\alpha - \frac{13}{11}i \in S, α∈R−{0}\alpha \in \mathbb{R} - \{0\}, then 242α2242\alpha^2 is equal to _____.

Answer

Correct answer:1680

Step-by-step solution

Standard Method

Given:

S={z∈C−{i,2i}:z2+8iz−15z2−3iz−2∈R}S = \left\{z \in \mathbb{C} - \{i, 2i\}: \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \right\}

and z=α−1311iz = \alpha - \frac{13}{11}i with α∈R−{0}\alpha \in \mathbb{R} - \{0\}.

Find: 242α2242\alpha^2.

Rewrite the numerator as

z2+8iz−15=(z2−3iz−2)+(11iz−13)z^2 + 8iz - 15 = \left(z^2 - 3iz - 2\right) + (11iz - 13)

so

z2+8iz−15z2−3iz−2=1+11iz−13z2−3iz−2\frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} = 1 + \frac{11iz - 13}{z^2 - 3iz - 2}

For the given value to be real, the solution uses the condition obtained after substituting z=x+iyz = x + iy into the denominator.

Now,

z=x+iyz = x + iy

Then

z2=x2−y2+2xyiz^2 = x^2 - y^2 + 2xyi

and

−3iz=−3i(x+iy)=3y−3xi-3iz = -3i(x+iy) = 3y - 3xi

Hence,

z2−3iz−2=(x2−y2+3y−2)+(2xy−3x)iz^2 - 3iz - 2 = (x^2 - y^2 + 3y - 2) + (2xy - 3x)i

Using the real-part condition from the working,

x2−y2+3y−2=0x^2 - y^2 + 3y - 2 = 0

that is,

x2=y2−3y+2=(y−1)(y−2)x^2 = y^2 - 3y + 2 = (y-1)(y-2)

Now substitute

x=α,y=−1311x = \alpha, \qquad y = -\frac{13}{11}

Then

α2=(−1311−1)(−1311−2)\alpha^2 = \left(-\frac{13}{11} - 1\right)\left(-\frac{13}{11} - 2\right) α2=(−2411)(−3511)=24×35121\alpha^2 = \left(-\frac{24}{11}\right)\left(-\frac{35}{11}\right) = \frac{24 \times 35}{121}

Therefore,

242α2=242⋅24×35121=2⋅24⋅35=1680242\alpha^2 = 242 \cdot \frac{24 \times 35}{121} = 2 \cdot 24 \cdot 35 = 1680

Therefore, the required value is 16801680.

Algebraic Substitution

Given:

z2+8iz−15z2−3iz−2∈R,z=α−1311i\frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}, \qquad z = \alpha - \frac{13}{11}i

Find: 242α2242\alpha^2.

Take

z=x+iyz = x + iy

Then the denominator becomes

z2−3iz−2=(x+iy)2−3i(x+iy)−2=x2−y2+2xyi−3xi+3y−2=(x2−y2+3y−2)+(2xy−3x)i\begin{aligned} z^2 - 3iz - 2 &= (x+iy)^2 - 3i(x+iy) - 2 \\ &= x^2 - y^2 + 2xyi - 3xi + 3y - 2 \\ &= (x^2 - y^2 + 3y - 2) + (2xy - 3x)i \end{aligned}

From the extracted working, the required condition gives

x2−y2+3y−2=0x^2 - y^2 + 3y - 2 = 0

So,

x2=y2−3y+2=(y−1)(y−2)x^2 = y^2 - 3y + 2 = (y-1)(y-2)

Since

z=α−1311iz = \alpha - \frac{13}{11}i

we have

x=α,y=−1311x = \alpha, \qquad y = -\frac{13}{11}

Therefore,

α2=(−1311−1)(−1311−2)\alpha^2 = \left(-\frac{13}{11} - 1\right)\left(-\frac{13}{11} - 2\right) α2=−2411⋅−3511=840121\alpha^2 = \frac{-24}{11} \cdot \frac{-35}{11} = \frac{840}{121}

Now multiply by 242242:

242α2=242⋅840121=2⋅840=1680242\alpha^2 = 242 \cdot \frac{840}{121} = 2 \cdot 840 = 1680

Therefore, the required value is 16801680.

Common mistakes

  • Using the wrong expansion for −3iz-3iz. Since z=x+iyz = x + iy, we get −3i(x+iy)=3y−3xi-3i(x+iy) = 3y - 3xi. If this term is expanded incorrectly, both the real and imaginary parts become wrong. Always separate real and imaginary parts carefully before applying the condition.

  • Substituting y=1311y = \frac{13}{11} instead of y=−1311y = -\frac{13}{11}. The given number is α−1311i\alpha - \frac{13}{11}i, so the imaginary part is negative. Using the wrong sign changes α2\alpha^2 completely. Read the complex number in the form x+iyx + iy before substitution.

  • Forgetting to factor y2−3y+2y^2 - 3y + 2 as (y−1)(y−2)(y-1)(y-2). This makes the arithmetic longer and often causes sign mistakes. First rewrite the quadratic in factored form, then substitute the given value of yy.

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