If z 1 and z 2 are two non-zero complex such that - z 1- z 2- z 1- z 2- then z 1- z 2 is equal toA- - B- -C- -2D- 0

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Integration Using Substitution

If z 1 and z ...

Question

If z₁ and z₂ are two non-zero complex such that $∣ z_{1} + z_{2} ∣=∣ z_{1} ∣ + ∣ z_{2} ∣,$ then arg $(z_{1}) -$ arg $(z_{2})$ is equal to

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Solution

The correct option is D $0$
CONVENTIONAL APPROACH

Let $z i = r_{1} (c o s θ_{1} + i s i n θ_{1}), z_{2} = r_{2} (c o s θ_{2} + i s i n θ_{2})$

$∴∣ z_{1} + z_{2} ∣ [(r_{1} c o s θ_{1} + r_{2} c o s θ_{2})^{2} + (r_{2} s i n θ_{1} + r_{2} s i n θ_{2})^{2}]^{\frac{1}{2}}$

$= [(r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} c o s (θ_{1} - θ_{2})^{2}]^{\frac{1}{2}} (∵ ∣ z_{1} + z_{2} ∣=∣ z_{1} ∣ + ∣ z_{2} ∣)$

Therefore $c o s (θ_{1} - θ_{2}) = 1 \Rightarrow θ_{1} - θ_{2} = 0 \Rightarrow θ_{1} = θ_{2}$

Thus arg $(z_{1}) - a r g (z_{2}) = 0.$

Tricks: $∣ z_{1} + z_{2} ∣=∣ z_{1} ∣ + ∣ z_{2} ∣ \Rightarrow z_{1}, z_{2}$ lies on same straight line.

$∴ a r g z_{1} = a r g z_{2} \Rightarrow a r g z_{1} - a r g z_{2} = 0$

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Hint: Square on both sides.

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If z1 and z2 are two non-zero complex such that ∣ z1+z2∣=∣ z1∣+∣ z2∣, then arg (z1)− arg (z2) is equal to

If z₁ and z₂ are two non-zero complex such that $∣ z_{1} + z_{2} ∣=∣ z_{1} ∣ + ∣ z_{2} ∣,$ then arg $(z_{1}) -$ arg $(z_{2})$ is equal to