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Complex Numbers question

2010 · Shift 1 · Q32
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  5. /2010 · Shift 1 · Q32

Complex Numbers question

2010 · Shift 1 · Q32

JEE AdvancedMathematicsComplex NumbersMultiple correct+4 / −1
Let z1{{z_1}}z1​ and z2{{z_2}}z2​ be two distinct complex number and let z =( 1 - t) z1{{z_1}}z1​ + t z2{{z_2}}z2​ for some real number t with 0 < t < 1. IfArg (w) denote the principal argument of a non-zero complex number w, then
  1. A
    ∣z−z1∣+∣z−z2∣=∣z1−z2∣\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = \left| {{z_1} - {z_2}} \right|∣z−z1​∣+∣z−z2​∣=∣z1​−z2​∣
  2. B
    Arg (z−z1)(z - {z_1})(z−z1​) = Arg (z−z2)(z - {z_2})(z−z2​)
  3. C
    ∣z−z1z‾−z‾1z2−z1z‾2−z‾1∣\left| {\begin{matrix} {z - {z_1}} & {\overline z - {{\overline z }_1}} \\ {{z_2} - {z_1}} & {{{\overline z }_2} - {{\overline z }_1}} \\ \end{matrix} } \right|​z−z1​z2​−z1​​z−z1​z2​−z1​​​ = 0
  4. D
    Arg (z−z1)(z - {z_1})(z−z1​) = Arg (z2−z1)({z_2} - {z_1})(z2​−z1​)
View written solutionFree

Correct answer: D, A, C

  1. Given expression for zzz

    \quad 0<t<1,$$ where $z_1,z_2$ are distinct complex numbers. Rewrite the differences: $$z-z_1=(1-t)z_1+t z_2-z_1=t(z_2-z_1),$$ $$z-z_2=(1-t)z_1+t z_2-z_2=-(1-t)(z_2-z_1)=(1-t)(z_1-z_2).$$ So $z$ lies on the line segment joining $z_1$ and $z_2$.

  2. Check Option A

    We have

    ∣z−z1∣=∣t(z2−z1)∣=t∣z2−z1∣,|z-z_1|=|t(z_2-z_1)|=t|z_2-z_1|,∣z−z1​∣=∣t(z2​−z1​)∣=t∣z2​−z1​∣, ∣z−z2∣=∣(1−t)(z1−z2)∣=(1−t)∣z1−z2∣.|z-z_2|=|(1-t)(z_1-z_2)|=(1-t)|z_1-z_2|.∣z−z2​∣=∣(1−t)(z1​−z2​)∣=(1−t)∣z1​−z2​∣.

    Therefore,

    ∣z−z1∣+∣z−z2∣=t∣z1−z2∣+(1−t)∣z1−z2∣=∣z1−z2∣.|z-z_1|+|z-z_2|=t|z_1-z_2|+(1-t)|z_1-z_2|=|z_1-z_2|.∣z−z1​∣+∣z−z2​∣=t∣z1​−z2​∣+(1−t)∣z1​−z2​∣=∣z1​−z2​∣.

    Hence A is correct.

  3. Check Option B

    From above,

    z−z1=t(z2−z1),z-z_1=t(z_2-z_1),z−z1​=t(z2​−z1​), z−z2=−(1−t)(z2−z1).z-z_2=-(1-t)(z_2-z_1).z−z2​=−(1−t)(z2​−z1​).

    Since t>0t>0t>0 and 1−t>01-t>01−t>0, multiplying by a positive real number does not change argument, but multiplying by −1-1−1 changes argument by π\piπ (mod 2π2\pi2π).

    Thus,

    \Arg(z−z2)=\Arg(z2−z1)+π(mod2π),\Arg(z-z_2)=\Arg(z_2-z_1)+\pi \pmod{2\pi},\Arg(z−z2​)=\Arg(z2​−z1​)+π(mod2π),

    so generally

    \Arg(z−z1)≠\Arg(z−z2).\Arg(z-z_1)\ne \Arg(z-z_2).\Arg(z−z1​)=\Arg(z−z2​).

    Hence B is incorrect.

  4. Check Option C

    The determinant is

    z-z_1 & \bar z-\bar z_1\\ z_2-z_1 & \bar z_2-\bar z_1 \end{vmatrix}.$$ Now, $$z-z_1=t(z_2-z_1),$$ $$\bar z-\bar z_1=t(\bar z_2-\bar z_1).$$ So the first row is $t$ times the second row. Therefore the determinant is zero. Hence **C is correct**.

  5. Check Option D

    Since

    z−z1=t(z2−z1),z-z_1=t(z_2-z_1),z−z1​=t(z2​−z1​),

    and t>0t>0t>0, multiplication by a positive real number does not change principal argument. Therefore,

    \Arg(z−z1)=\Arg(z2−z1).\Arg(z-z_1)=\Arg(z_2-z_1).\Arg(z−z1​)=\Arg(z2​−z1​).

    Hence D is correct.

  6. Final answer

    Correct options are:

    A, C, D\boxed{A,\ C,\ D}A, C, D​

  7. Comparison with stored answer

    Stored correct answer: D,A,CD, A, CD,A,C

    This matches exactly (order does not matter).

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