Question

Let \( z_{3} z_{2} \) be two complex numbers such that \( \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \). Then

(a) \( \arg \left(z_{1}\right) \neq \arg \left(z_{2}\right) \)

(b) \( \arg \left( z _{1}\right)+\arg \left( z _{2}\right)=0 \)

(c) \( \left.\cos \frac{z_{1}}{z_{z}}\right) =0 \)

(d) None of these

Answer 1

\(\because\) z\(\bar z\) = |z|²

\(\therefore\) |z₁ + z₂|2 = (z₁ + z₂) (\(\bar z_1+\bar z_2\)) (\(\because \overline{z_1+z_2}=\bar z_1 + \bar z_2\))

 = z₁ \(\bar z_1\) + z₁\(\bar z_2\) + z₂\(\bar z_1\) + z₂\(\bar z_2\)

 = |z₁|² + |z₂|² + z₁ \(\bar z_2\) + \(\overline{z_1\bar z_2}\)

 = |z₁|² + |z₂|² + z₁ \(\bar z_2\) + 2Re (z₁, \(\bar z_2\))

(\(\because z + \bar z=2Re(z)\))

Now, (|z₁ + z₂|)² = |z₁|² + |z₂|² + 2|z₁||z₂|

⇒ Re(z₁\(\bar z_2\))  = |z₁||z₂|---(1)

Let z₁ = a + ib

z₂ = c + id

\(\therefore\) |z₁| = \(\sqrt{a^2+b^2}\)

|z₂| = \(\sqrt{c^2+d^2}\) 

z₁ \(\bar z_2\)  = (a + ib) (c - id)

 = ac + bd + i(bc - ad)

Re(z₁ \(\bar z_2\)) = ac + bd.

From equation (1), we get

ac + bd = \(\sqrt{a^2+b^2}\sqrt{c^2+d^2}\) 

⇒ a²c² + b²d² + 2abcd = (a² + b²)(c² + d²)

(By squaring on both sides)

⇒ a²c² + b²d² + 2abcd = a²c² + a²d² + b²c² + b²d²

⇒ (ad - bc)² = 0

⇒ ad = bc

⇒ b/a = d/c

tan θ₁ = tan θ₂ where θ₁ = arg(z₁), θ₂ = arg(z₂)

⇒ θ₁ = θ₂ +nπ, n\(\in\) z