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Complex Numbers question

2005 · Shift 0 · Q94
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Complex Numbers question

2005 · Shift 0 · Q94

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z1{z_1}z1​ and z2{z_2}z2​ are two non-zero complex numbers such that  ∣z1+z2∣=∣z1∣+∣z2∣\,\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|∣z1​+z2​∣=∣z1​∣+∣z2​∣, then arg z1{z_1}z1​- arg z2{z_2}z2​ is equal to :
  1. A
    π2 {\pi \over 2}\,2π​
  2. B
    −π- \pi−π
  3. C
    0
  4. D
    −π2{{ - \pi } \over 2}2−π​
View written solutionFree

Correct answer: C

  1. We use the equality condition for the triangle inequality:

∣z1+z2∣≤∣z1∣+∣z2∣|z_1+z_2| \le |z_1|+|z_2|∣z1​+z2​∣≤∣z1​∣+∣z2​∣

and equality holds iff z1z_1z1​ and z2z_2z2​ have the same direction in the complex plane, i.e. one is a non-negative real multiple of the other.

  1. Let us verify this algebraically by squaring both sides:

∣z1+z2∣2=(∣z1∣+∣z2∣)2|z_1+z_2|^2=(|z_1|+|z_2|)^2∣z1​+z2​∣2=(∣z1​∣+∣z2​∣)2

Now,

∣z1+z2∣2=∣z1∣2+∣z2∣2+2Re⁡(z1z2‾)|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\operatorname{Re}(z_1\overline{z_2})∣z1​+z2​∣2=∣z1​∣2+∣z2​∣2+2Re(z1​z2​​)

and

(∣z1∣+∣z2∣)2=∣z1∣2+∣z2∣2+2∣z1∣∣z2∣(|z_1|+|z_2|)^2=|z_1|^2+|z_2|^2+2|z_1||z_2|(∣z1​∣+∣z2​∣)2=∣z1​∣2+∣z2​∣2+2∣z1​∣∣z2​∣

So,

Re⁡(z1z2‾)=∣z1∣∣z2∣\operatorname{Re}(z_1\overline{z_2})=|z_1||z_2|Re(z1​z2​​)=∣z1​∣∣z2​∣

  1. Write

z1=∣z1∣eiθ1,z2=∣z2∣eiθ2z_1=|z_1|e^{i\theta_1}, \qquad z_2=|z_2|e^{i\theta_2}z1​=∣z1​∣eiθ1​,z2​=∣z2​∣eiθ2​

Then

z1z2‾=∣z1∣∣z2∣ei(θ1−θ2)z_1\overline{z_2}=|z_1||z_2|e^{i(\theta_1-\theta_2)}z1​z2​​=∣z1​∣∣z2​∣ei(θ1​−θ2​)

Hence,

Re⁡(z1z2‾)=∣z1∣∣z2∣cos⁡(θ1−θ2)\operatorname{Re}(z_1\overline{z_2})=|z_1||z_2|\cos(\theta_1-\theta_2)Re(z1​z2​​)=∣z1​∣∣z2​∣cos(θ1​−θ2​)

Using the equality obtained above,

∣z1∣∣z2∣cos⁡(θ1−θ2)=∣z1∣∣z2∣|z_1||z_2|\cos(\theta_1-\theta_2)=|z_1||z_2|∣z1​∣∣z2​∣cos(θ1​−θ2​)=∣z1​∣∣z2​∣

Since z1,z2≠0z_1,z_2\neq 0z1​,z2​=0, we have ∣z1∣∣z2∣≠0|z_1||z_2|\neq 0∣z1​∣∣z2​∣=0. Therefore,

cos⁡(θ1−θ2)=1\cos(\theta_1-\theta_2)=1cos(θ1​−θ2​)=1

Thus,

θ1−θ2=2nπ,n∈Z\theta_1-\theta_2=2n\pi, \quad n\in \mathbb{Z}θ1​−θ2​=2nπ,n∈Z

So the principal value among the given options is

arg⁡z1−arg⁡z2=0\arg z_1-\arg z_2=0argz1​−argz2​=0

  1. Check options:
  • A: π2\dfrac{\pi}{2}2π​ — incorrect
  • B: −π-\pi−π — incorrect
  • C: 000 — correct
  • D: −π2-\dfrac{\pi}{2}−2π​ — incorrect

Therefore, the correct answer is C.

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