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Complex Numbers question

2022 · Shift 1 · Q22
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Complex Numbers question

2022 · Shift 1 · Q22

JEE AdvancedMathematicsComplex NumbersNumerical+3 / −1
Let zzz be a complex number with a non-zero imaginary part. If 2+3z+4z22−3z+4z2\frac{2+3 z+4 z^{2}}{2-3 z+4 z^{2}}2−3z+4z22+3z+4z2​ is a real number, then the value of ∣z∣2|z|^{2}∣z∣2 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 0.49TO0.51

Let

R=2+3z+4z22−3z+4z2R= \frac{2+3z+4z^2}{2-3z+4z^2}R=2−3z+4z22+3z+4z2​

and we are given that RRR is real, with Im⁡(z)≠0\operatorname{Im}(z)\neq 0Im(z)=0.

We need to find ∣z∣2|z|^2∣z∣2.

1. Write zzz in terms of real and imaginary parts

Let

z=x+iy,z=x+iy,z=x+iy,

where x,y∈Rx,y\in\mathbb Rx,y∈R and y≠0y\neq 0y=0.

Then

z2=(x+iy)2=x2−y2+2ixy.z^2=(x+iy)^2=x^2-y^2+2ixy.z2=(x+iy)2=x2−y2+2ixy.

So, [ 2+3z+4z^2 =2+3(x+iy)+4(x^2-y^2+2ixy). ] Hence its real and imaginary parts are

A=2+3x+4x2−4y2,A=2+3x+4x^2-4y^2,A=2+3x+4x2−4y2, B=3y+8xy=y(3+8x).B=3y+8xy=y(3+8x).B=3y+8xy=y(3+8x).

Thus,

2+3z+4z2=A+iB.2+3z+4z^2=A+iB.2+3z+4z2=A+iB.

Similarly, [ 2-3z+4z^2 =2-3(x+iy)+4(x^2-y^2+2ixy), ] so its real and imaginary parts are

C=2−3x+4x2−4y2,C=2-3x+4x^2-4y^2,C=2−3x+4x2−4y2, D=−3y+8xy=y(−3+8x).D=-3y+8xy=y(-3+8x).D=−3y+8xy=y(−3+8x).

Thus,

2−3z+4z2=C+iD.2-3z+4z^2=C+iD.2−3z+4z2=C+iD.

2. Condition for the quotient to be real

For

A+iBC+iD\frac{A+iB}{C+iD}C+iDA+iB​

to be real, we need

BC−AD=0.BC-AD=0.BC−AD=0.

So,

BC−AD=0.B C-A D=0.BC−AD=0.

Substitute the values:

y(3+8x)(2−3x+4x2−4y2)−(2+3x+4x2−4y2) y(−3+8x)=0.y(3+8x)(2-3x+4x^2-4y^2)- (2+3x+4x^2-4y^2)\,y(-3+8x)=0.y(3+8x)(2−3x+4x2−4y2)−(2+3x+4x2−4y2)y(−3+8x)=0.

Since y≠0y\neq 0y=0, divide by yyy:

(3+8x)(2−3x+4x2−4y2)−(2+3x+4x2−4y2)(−3+8x)=0.(3+8x)(2-3x+4x^2-4y^2)-(2+3x+4x^2-4y^2)(-3+8x)=0.(3+8x)(2−3x+4x2−4y2)−(2+3x+4x2−4y2)(−3+8x)=0.

Rewrite as

(3+8x)(2−3x+4x2−4y2)+(3−8x)(2+3x+4x2−4y2)=0.(3+8x)(2-3x+4x^2-4y^2)+(3-8x)(2+3x+4x^2-4y^2)=0.(3+8x)(2−3x+4x2−4y2)+(3−8x)(2+3x+4x2−4y2)=0.

Now expand:

First term:

(3+8x)(2−3x+4x2−4y2)=6+7x−12x2+32x3−12y2−32xy2.(3+8x)(2-3x+4x^2-4y^2) =6+7x-12x^2+32x^3-12y^2-32xy^2.(3+8x)(2−3x+4x2−4y2)=6+7x−12x2+32x3−12y2−32xy2.

Second term:

(3−8x)(2+3x+4x2−4y2)=6−7x−12x2−32x3−12y2+32xy2.(3-8x)(2+3x+4x^2-4y^2) =6-7x-12x^2-32x^3-12y^2+32xy^2.(3−8x)(2+3x+4x2−4y2)=6−7x−12x2−32x3−12y2+32xy2.

Adding,

12−24x2−24y2=0.12-24x^2-24y^2=0.12−24x2−24y2=0.

So,

1−2x2−2y2=01-2x^2-2y^2=01−2x2−2y2=0

which gives

x2+y2=12.x^2+y^2=\frac12.x2+y2=21​.

But

∣z∣2=x2+y2.|z|^2=x^2+y^2.∣z∣2=x2+y2.

Therefore,

∣z∣2=12.|z|^2=\frac12.∣z∣2=21​.

3. Final answer

∣z∣2=12|z|^2=\frac12∣z∣2=21​

So the required integer-type/numerical value is

0.5\boxed{0.5}0.5​

4. Comparison with stored answer

Stored correct answer: 0.490.490.49 to 0.510.510.51.

Our derived answer is 0.50.50.5, which lies in this range. Hence the answer agrees with the stored answer.

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