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

If ∣z_1 −z_2 ∣ = ∣z_1 ∣+∣z_2 ∣ , then prove that arg((z_1 /z_2 ))=π .

$${If}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid\:,\:{then}\:{prove}\: \\ $$$${that}\:{arg}\left(\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\right)=\pi\:. \\ $$

Answered by MJS last updated on 04/Dec/18

z_1 =pe^(iα) ; z_2 =qe^(iβ) ∣z_1 −z_2 ∣=(√(p^2 +q^2 −2pq cos (α−β))) ∣z_1 ∣+∣z_2 ∣=∣p∣+∣q∣ ⇒ −2pqcos (α−β)=2pq ⇒ cos (α−β)=−1 ⇒ α−β=π [(z_1 /z_2 )=(p/q)e^(i(α−β)) ]

$${z}_{\mathrm{1}} ={p}\mathrm{e}^{\mathrm{i}\alpha} ;\:\mathrm{z}_{\mathrm{2}} ={q}\mathrm{e}^{\mathrm{i}\beta} \\ $$$$\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pq}\:\mathrm{cos}\:\left(\alpha−\beta\right)} \\ $$$$\mid{z}_{\mathrm{1}} \mid+\mid{z}_{\mathrm{2}} \mid=\mid{p}\mid+\mid{q}\mid \\ $$$$\Rightarrow\:−\mathrm{2}{pq}\mathrm{cos}\:\left(\alpha−\beta\right)=\mathrm{2}{pq} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left(\alpha−\beta\right)=−\mathrm{1}\:\Rightarrow\:\alpha−\beta=\pi \\ $$$$\left[\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }=\frac{{p}}{{q}}\mathrm{e}^{\mathrm{i}\left(\alpha−\beta\right)} \right] \\ $$

Commented by MJS last updated on 04/Dec/18

$$…\mathrm{corrected} \\ $$

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

pls explain 2^(nd) line .^.^.^(...^(....) )

$${pls}\:{explain}\:\mathrm{2}^{{nd}} \:{line}\:\overset{\overset{\overset{..\overset{….} {.}} {.}} {.}} {.} \\ $$

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 =pcos α −qcos β +i(psin α −qsin β) ∣z_1 −z_2 ∣=(√((pcos α −qcos β)^2 +(psin α −qsin β)^2 ))= =(√(p^2 +q^2 −2pq(cos α cos β +sin α sin β)))= =(√(p^2 +q^2 −2pq cos (α−β)))

$${z}_{\mathrm{1}} ={p}\left(\mathrm{cos}\:\alpha\:+\mathrm{isin}\:\alpha\right) \\ $$$${z}_{\mathrm{2}} ={q}\left(\mathrm{cos}\:\beta\:+\mathrm{isin}\:\beta\right) \\ $$$${z}_{\mathrm{1}} −{z}_{\mathrm{2}} ={p}\mathrm{cos}\:\alpha\:−{q}\mathrm{cos}\:\beta\:+\mathrm{i}\left({p}\mathrm{sin}\:\alpha\:−{q}\mathrm{sin}\:\beta\right) \\ $$$$\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid=\sqrt{\left({p}\mathrm{cos}\:\alpha\:−{q}\mathrm{cos}\:\beta\right)^{\mathrm{2}} +\left({p}\mathrm{sin}\:\alpha\:−{q}\mathrm{sin}\:\beta\right)^{\mathrm{2}} }= \\ $$$$=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pq}\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:+\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\right)}= \\ $$$$=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{pq}\:\mathrm{cos}\:\left(\alpha−\beta\right)} \\ $$

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

thank you sir! ��

Commented by MJS last updated on 04/Dec/18

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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