If Z 1 and z 2 are two complex numbers satisfying the equation - z 1- z 2- z 1- z 2-1- then z 1- z 2 is a number which isA- Zero or purely imaginaryB- None of theseC- Negative realD- Positive real

Given | z_1 + z_2/z_1 z_2 |=1 ⇒z_1 + z_2/z_1 z_2 = cos θ + i sin θ (say) ⇒z_1/z_2 = 1 + cos θ + i sin θ/ 1 + cos θ + i sin θ = i cot θ/2 which is zero, i ...

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Byju's Answer

Standard XII

Mathematics

Purely Real

If Z 1 and z ...

Question

If $z_{1}$ and $z_{2}$ are two complex numbers satisfying the equation $∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ = 1,$ then $\frac{z_{1}}{z_{2}}$ is a number which is

Positive real

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Negative real

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Zero or purely imaginary

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None of these

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Solution

The correct option is B
Negative real

Given $∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ = 1 \Rightarrow \frac{z_{1} + z_{2}}{z_{1} - z_{2}} = c o s θ + i s i n θ$ (say)
$\Rightarrow \frac{z_{1}}{z_{2}} = \frac{1 + c o s θ + i s i n θ}{- 1 + c o s θ + i s i n θ} = i c o t \frac{θ}{2}$
which is zero, if $θ = n π (n ϵ 1),$ and is otherwise purely imaginary.

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If z1 and z2 are two complex numbers satisfying the equation ∣∣z1+z2z1−z2∣∣=1, then z1z2 is a number which is

The correct option is B Negative real Given ∣∣z1+z2z1−z2∣∣=1⇒z1+z2z1−z2=cos θ+i sin θ (say) ⇒z1z2=1+cos θ+i sin θ−1+cos θ+i sin θ=icotθ2 which is zero, if θ=nπ(nϵ1), and is otherwise purely imaginary.

If $z_{1}$ and $z_{2}$ are two complex numbers satisfying the equation $∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ = 1,$ then $\frac{z_{1}}{z_{2}}$ is a number which is