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Question Number 19352 by Tinkutara last updated on 10/Aug/17

Prove that ∣z_1 − z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 − 2∣z_1 ∣ ∣z_2 ∣ cos (θ_1 − θ_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$−\:\mathrm{2}\mid{z}_{\mathrm{1}} \mid\:\mid{z}_{\mathrm{2}} \mid\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \:−\:\theta_{\mathrm{2}} \right) \\ $$

Answered by ajfour last updated on 10/Aug/17

(r_1 cos θ_1 −r_2 cos θ_2 )^2 +(r_1 sin θ_1 −r_2 sin θ_2 )^2 =r_1 ^2 +r_2 ^2 −2r_1 r_2 (cos θ_1 cos θ_2 +sin θ_1 sin θ_2 ) =∣z_1 ∣^2 +∣z_2 ∣^2 −2∣z_1 ∣∣z_2 ∣cos (θ_1 −θ_2 ) .

$$\left(\mathrm{r}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} −\mathrm{r}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{r}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} −\mathrm{r}_{\mathrm{2}} \mathrm{sin}\:\theta_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$=\mathrm{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{r}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2r}_{\mathrm{1}} \mathrm{r}_{\mathrm{2}} \left(\mathrm{cos}\:\theta_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{2}} +\mathrm{sin}\:\theta_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{2}} \right) \\ $$$$=\mid\mathrm{z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} −\mathrm{2}\mid\mathrm{z}_{\mathrm{1}} \mid\mid\mathrm{z}_{\mathrm{2}} \mid\mathrm{cos}\:\left(\theta_{\mathrm{1}} −\theta_{\mathrm{2}} \right)\:. \\ $$

Commented by Tinkutara last updated on 10/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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