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Complex Numbers question

2024 · 5 Apr · Shift 1 · Q35
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  5. /2024 · 5 Apr · Shift 1 · Q35

Complex Numbers question

2024 · 5 Apr · Shift 1 · Q35

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Consider the following two statements : Statement I: For any two non-zero complex numbers z1,z2,(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣), and z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }z1​,z2​,(∣z1​∣+∣z2​∣)​∣z1​∣z1​​+∣z2​∣z2​​​≤2(∣z1​∣+∣z2​∣), and  Statement II : If x,y,zx, y, zx,y,z are three distinct complex numbers and a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c}a,b,c are three positive real numbers such that a∣y−z∣=b∣z−x∣=c∣x−y∣\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}∣y−z∣a​=∣z−x∣b​=∣x−y∣c​, then a2y−z+b2z−x+c2x−y=1\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1y−za2​+z−xb2​+x−yc2​=1. Between the above two statements,
  1. A
    both Statement I and Statement II are incorrect.
  2. B
    Statement I is correct but Statement II is incorrect.
  3. C
    Statement I is incorrect but Statement II is correct.
  4. D
    both Statement I and Statement II are correct.
View written solutionFree

Correct answer: B

  1. Statement I

We need to check whether

(∣z1∣+∣z2∣)∣z1∣z1∣+z2∣z2∣∣≤2(∣z1∣+∣z2∣)(|z_1|+|z_2|)\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right| \le 2(|z_1|+|z_2|)(∣z1​∣+∣z2​∣)​∣z1​∣z1​​+∣z2​∣z2​​​≤2(∣z1​∣+∣z2​∣)

for any non-zero complex numbers z1,z2z_1,z_2z1​,z2​.

Since ∣z1∣+∣z2∣>0|z_1|+|z_2|>0∣z1​∣+∣z2​∣>0, divide both sides by this positive quantity:

∣z1∣z1∣+z2∣z2∣∣≤2.\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right| \le 2.​∣z1​∣z1​​+∣z2​∣z2​​​≤2.

Now,

∣z1∣z1∣∣=1,∣z2∣z2∣∣=1.\left|\frac{z_1}{|z_1|}\right|=1, \qquad \left|\frac{z_2}{|z_2|}\right|=1.​∣z1​∣z1​​​=1,​∣z2​∣z2​​​=1.

So by triangle inequality,

∣z1∣z1∣+z2∣z2∣∣≤∣z1∣z1∣∣+∣z2∣z2∣∣=1+1=2.\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right| \le \left|\frac{z_1}{|z_1|}\right|+\left|\frac{z_2}{|z_2|}\right|=1+1=2.​∣z1​∣z1​​+∣z2​∣z2​​​≤​∣z1​∣z1​​​+​∣z2​∣z2​​​=1+1=2.

Hence Statement I is correct.

  1. Statement II

Given

a∣y−z∣=b∣z−x∣=c∣x−y∣.\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}.∣y−z∣a​=∣z−x∣b​=∣x−y∣c​.

Let the common value be k>0k>0k>0. Then

a=k∣y−z∣,b=k∣z−x∣,c=k∣x−y∣.a=k|y-z|,\quad b=k|z-x|,\quad c=k|x-y|.a=k∣y−z∣,b=k∣z−x∣,c=k∣x−y∣.

Therefore,

a2y−z+b2z−x+c2x−y=k2(∣y−z∣2y−z+∣z−x∣2z−x+∣x−y∣2x−y).\frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y} = k^2\left(\frac{|y-z|^2}{y-z}+\frac{|z-x|^2}{z-x}+\frac{|x-y|^2}{x-y}\right).y−za2​+z−xb2​+x−yc2​=k2(y−z∣y−z∣2​+z−x∣z−x∣2​+x−y∣x−y∣2​).

Use the identity

∣w∣2w=wˉ(w≠0),\frac{|w|^2}{w}=\bar w \qquad (w\ne 0),w∣w∣2​=wˉ(w=0),

since ∣w∣2=wwˉ|w|^2=w\bar w∣w∣2=wwˉ. Thus,

∣y−z∣2y−z=(y−z)‾=yˉ−zˉ,\frac{|y-z|^2}{y-z}=\overline{(y-z)}=\bar y-\bar z,y−z∣y−z∣2​=(y−z)​=yˉ​−zˉ, ∣z−x∣2z−x=(z−x)‾=zˉ−xˉ,\frac{|z-x|^2}{z-x}=\overline{(z-x)}=\bar z-\bar x,z−x∣z−x∣2​=(z−x)​=zˉ−xˉ, ∣x−y∣2x−y=(x−y)‾=xˉ−yˉ.\frac{|x-y|^2}{x-y}=\overline{(x-y)}=\bar x-\bar y.x−y∣x−y∣2​=(x−y)​=xˉ−yˉ​.

Adding,

(yˉ−zˉ)+(zˉ−xˉ)+(xˉ−yˉ)=0.(\bar y-\bar z)+(\bar z-\bar x)+(\bar x-\bar y)=0.(yˉ​−zˉ)+(zˉ−xˉ)+(xˉ−yˉ​)=0.

Hence

a2y−z+b2z−x+c2x−y=k2⋅0=0.\frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=k^2\cdot 0=0.y−za2​+z−xb2​+x−yc2​=k2⋅0=0.

But the statement claims this sum equals 111, which is false.

So Statement II is incorrect.

  1. Conclusion
  • Statement I: correct
  • Statement II: incorrect

Therefore the correct option is

B\boxed{\text{B}}B​
  1. Comparison with stored answer

Stored correct answer: B

Our derived answer matches it.

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