If-z-1-z-2-z-1-z-2-then-prove-that-arg-z-1-z-2-pi-

Question Number 49177 by rahul 19 last updated on 04/Dec/18

I f ∣ z_{1} - z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣, t h e n p r o v e

t h a t a r g (\frac{z_{1}}{z_{2}}) = π .

Answered by MJS last updated on 04/Dec/18

z_{1} = p e^{i α}; z_{2} = q e^{i β}

∣ z_{1} - z_{2} ∣= \sqrt{p^{2} + q^{2} - 2 p q \cos (α - β)}

∣ z_{1} ∣ + ∣ z_{2} ∣=∣ p ∣ + ∣ q ∣

\Rightarrow - 2 p q \cos (α - β) = 2 p q

\Rightarrow \cos (α - β) = - 1 \Rightarrow α - β = π

[\frac{z_{1}}{z_{2}} = \frac{p}{q} e^{i (α - β)}]

Commented by MJS last updated on 04/Dec/18

\dots corrected

Commented by rahul 19 last updated on 04/Dec/18

p l s e x p l a i n 2^{n d} l i n e \overset{\overset{\overset{. . \overset{\dots .}{.}}{.}}{.}}{.}

Commented by MJS last updated on 04/Dec/18

z_{1} = p (\cos α + isin α)

z_{2} = q (\cos β + isin β)

z_{1} - z_{2} = p \cos α - q \cos β + i (p \sin α - q \sin β)

∣ z_{1} - z_{2} ∣= \sqrt{{(p \cos α - q \cos β)}^{2} + {(p \sin α - q \sin β)}^{2}} =

= \sqrt{p^{2} + q^{2} - 2 p q (\cos α \cos β + \sin α \sin β)} =

= \sqrt{p^{2} + q^{2} - 2 p q \cos (α - β)}

Commented by rahul 19 last updated on 04/Dec/18

thank you sir! ��

Commented by MJS last updated on 04/Dec/18

{you}^{'} re welcome

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