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Complex Numbers question

2023 · 11 Apr · Shift 2 · Q41
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Complex Numbers question

2023 · 11 Apr · Shift 2 · Q41

JEE MainMathematicsComplex NumbersNumerical+4 / −1
Let S={z∈C−{i,2i}:z2+8iz−15z2−3iz−2∈R}\mathrm{S}=\left\{z \in \mathbb{C}-\{i, 2 i\}: \frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \in \mathbb{R}\right\}S={z∈C−{i,2i}:z2−3iz−2z2+8iz−15​∈R}. If α−1311i∈S,α∈R−{0}\alpha-\frac{13}{11} i \in \mathrm{S}, \alpha \in \mathbb{R}-\{0\}α−1113​i∈S,α∈R−{0}, then 242α2242 \alpha^{2}242α2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1680

Let w= rac{z^{2}+8iz-15}{z^{2}-3iz-2}. We need those z∈C∖{i,2i}z\in\mathbb C\setminus\{i,2i\}z∈C∖{i,2i} for which w∈Rw\in\mathbb Rw∈R.

We are given that z=α−1311i,α∈R∖{0},z=\alpha-\frac{13}{11}i,\qquad \alpha\in\mathbb R\setminus\{0\},z=α−1113​i,α∈R∖{0}, and this z∈Sz\in Sz∈S. So we must impose that the above fraction is real.

1. Factor the numerator and denominator

Observe: z2+8iz−15=(z+3i)(z+5i),z^{2}+8iz-15=(z+3i)(z+5i),z2+8iz−15=(z+3i)(z+5i), since 3i+5i=8i3i+5i=8i3i+5i=8i and (3i)(5i)=−15(3i)(5i)=-15(3i)(5i)=−15.

Also, z2−3iz−2=(z−i)(z−2i),z^{2}-3iz-2=(z-i)(z-2i),z2−3iz−2=(z−i)(z−2i), since (−i)+(−2i)=−3i(-i)+(-2i)=-3i(−i)+(−2i)=−3i and (−i)(−2i)=−2(-i)(-2i)=-2(−i)(−2i)=−2.

Hence w=(z+3i)(z+5i)(z−i)(z−2i).w=\frac{(z+3i)(z+5i)}{(z-i)(z-2i)}.w=(z−i)(z−2i)(z+3i)(z+5i)​.

2. Substitute z=α−1311iz=\alpha-\frac{13}{11}iz=α−1113​i

Compute each factor: z+3i=α+(3−1311)i=α+2011i,z+3i=\alpha+\left(3-\frac{13}{11}\right)i=\alpha+\frac{20}{11}i,z+3i=α+(3−1113​)i=α+1120​i, z+5i=α+(5−1311)i=α+4211i,z+5i=\alpha+\left(5-\frac{13}{11}\right)i=\alpha+\frac{42}{11}i,z+5i=α+(5−1113​)i=α+1142​i, z−i=α−(1311+1)i=α−2411i,z-i=\alpha-\left(\frac{13}{11}+1\right)i=\alpha-\frac{24}{11}i,z−i=α−(1113​+1)i=α−1124​i, z−2i=α−(1311+2)i=α−3511i.z-2i=\alpha-\left(\frac{13}{11}+2\right)i=\alpha-\frac{35}{11}i.z−2i=α−(1113​+2)i=α−1135​i.

So w=(α+2011i)(α+4211i)(α−2411i)(α−3511i).w=\frac{\left(\alpha+\frac{20}{11}i\right)\left(\alpha+\frac{42}{11}i\right)}{\left(\alpha-\frac{24}{11}i\right)\left(\alpha-\frac{35}{11}i\right)}.w=(α−1124​i)(α−1135​i)(α+1120​i)(α+1142​i)​.

3. Expand numerator and denominator

Numerator: N=(α+2011i)(α+4211i)N=\left(\alpha+\frac{20}{11}i\right)\left(\alpha+\frac{42}{11}i\right)N=(α+1120​i)(α+1142​i) =α2+α(6211i)+840121i2=\alpha^2+\alpha\left(\frac{62}{11}i\right)+\frac{840}{121}i^2=α2+α(1162​i)+121840​i2 =(α2−840121)+62α11i.=\left(\alpha^2-\frac{840}{121}\right)+\frac{62\alpha}{11}i.=(α2−121840​)+1162α​i.

Denominator: D=(α−2411i)(α−3511i)D=\left(\alpha-\frac{24}{11}i\right)\left(\alpha-\frac{35}{11}i\right)D=(α−1124​i)(α−1135​i) =α2−α(5911i)+840121i2=\alpha^2-\alpha\left(\frac{59}{11}i\right)+\frac{840}{121}i^2=α2−α(1159​i)+121840​i2 =(α2−840121)−59α11i.=\left(\alpha^2-\frac{840}{121}\right)-\frac{59\alpha}{11}i.=(α2−121840​)−1159α​i.

Thus w=A+iBA−iC,w=\frac{A+iB}{A-iC},w=A−iCA+iB​, where A=α2−840121,B=62α11,C=59α11.A=\alpha^2-\frac{840}{121},\quad B=\frac{62\alpha}{11},\quad C=\frac{59\alpha}{11}.A=α2−121840​,B=1162α​,C=1159α​.

4. Condition for www to be real

For A+iBA−iC\frac{A+iB}{A-iC}A−iCA+iB​ to be real, multiply numerator and denominator by the conjugate of denominator: w=(A+iB)(A+iC)A2+C2.w=\frac{(A+iB)(A+iC)}{A^2+C^2}.w=A2+C2(A+iB)(A+iC)​.

The imaginary part of the numerator must vanish.

Now, (A+iB)(A+iC)=A2−BC+iA(B+C).(A+iB)(A+iC)=A^2-BC+iA(B+C).(A+iB)(A+iC)=A2−BC+iA(B+C).

So the imaginary part is zero iff A(B+C)=0.A(B+C)=0.A(B+C)=0.

Hence either

  1. A=0A=0A=0, or
  2. B+C=0B+C=0B+C=0.

But B+C=62α11+59α11=121α11=11α.B+C=\frac{62\alpha}{11}+\frac{59\alpha}{11}=\frac{121\alpha}{11}=11\alpha.B+C=1162α​+1159α​=11121α​=11α. Since α≠0\alpha\neq 0α=0, we have B+C≠0B+C\neq 0B+C=0.

Therefore necessarily A=0.A=0.A=0.

So α2−840121=0\alpha^2-\frac{840}{121}=0α2−121840​=0 ⇒α2=840121.\Rightarrow \alpha^2=\frac{840}{121}.⇒α2=121840​.

5. Compute 242α2242\alpha^2242α2

242α2=242⋅840121=2⋅840=1680.242\alpha^2=242\cdot \frac{840}{121}=2\cdot 840=1680.242α2=242⋅121840​=2⋅840=1680.

6. Final answer

1680\boxed{1680}1680​

This matches the stored correct answer.

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