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Question Number 19604 by ajfour last updated on 13/Aug/17

Commented by ajfour last updated on 13/Aug/17

solution to Q.19508 To prove p=∣z_1 −z_0 ∣=∣((α^� z_1 +αz_1 ^� +2c)/(2∣α∣))∣

$$\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{19508} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{0}} \mid=\mid\frac{\bar {\alpha}\mathrm{z}_{\mathrm{1}} +\alpha\bar {\mathrm{z}}_{\mathrm{1}} +\mathrm{2c}}{\mathrm{2}\mid\alpha\mid}\mid \\ $$

Answered by ajfour last updated on 13/Aug/17

z_F =α (see Q.19592) z_0 =z_1 −λα and z_0 ^� =z_1 ^� −λα^� as z_0 lies on line α^� z+αz^� +2c=0 we have α^� (z_1 −λα)+α(z_1 ^� −λα^� )+2c=0 or α^� z_1 +αz_1 ^� +2c=2λ∣α∣^2 ...(i) p=∣z_1 −z_0 ∣=∣λα∣ =∣λ∣∣α∣ using (i) we get: p=∣z_1 −z_0 ∣=∣((α^� z_1 +αz_1 ^� +2c)/(2∣α∣))∣.

$$\mathrm{z}_{\mathrm{F}} =\alpha\:\:\:\:\:\:\left(\mathrm{see}\:\mathrm{Q}.\mathrm{19592}\right) \\ $$$$\mathrm{z}_{\mathrm{0}} =\mathrm{z}_{\mathrm{1}} −\lambda\alpha\:\:\:\mathrm{and}\:\:\:\bar {\mathrm{z}}_{\mathrm{0}} =\bar {\mathrm{z}}_{\mathrm{1}} −\lambda\bar {\alpha} \\ $$$$\mathrm{as}\:\mathrm{z}_{\mathrm{0}} \:\mathrm{lies}\:\mathrm{on}\:\mathrm{line}\:\bar {\alpha}\mathrm{z}+\alpha\bar {\mathrm{z}}+\mathrm{2c}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\bar {\alpha}\left(\mathrm{z}_{\mathrm{1}} −\lambda\alpha\right)+\alpha\left(\bar {\mathrm{z}}_{\mathrm{1}} −\lambda\bar {\alpha}\right)+\mathrm{2c}=\mathrm{0} \\ $$$$\mathrm{or}\:\:\:\bar {\alpha}\mathrm{z}_{\mathrm{1}} +\alpha\bar {\mathrm{z}}_{\mathrm{1}} +\mathrm{2c}=\mathrm{2}\lambda\mid\alpha\mid^{\mathrm{2}} \:\:\:...\left(\mathrm{i}\right) \\ $$$$\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{0}} \mid=\mid\lambda\alpha\mid\:=\mid\lambda\mid\mid\alpha\mid \\ $$$$\mathrm{using}\:\left(\mathrm{i}\right)\:\mathrm{we}\:\mathrm{get}: \\ $$$$\:\:\:\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{0}} \mid=\mid\frac{\bar {\alpha}\mathrm{z}_{\mathrm{1}} +\alpha\bar {\mathrm{z}}_{\mathrm{1}} +\mathrm{2c}}{\mathrm{2}\mid\alpha\mid}\mid. \\ $$

Commented by Tinkutara last updated on 13/Aug/17

Thank you very much Sir!

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

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