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

Commented by ajfour last updated on 13/Aug/17

$$\mathrm{solution}\mathrm{to}\mathrm{Q}.19508$$$$\mathrm{To}\mathrm{prove}\mathrm{p}=\mid {\mathrm{z}}_1-{\mathrm{z}}_0\mid =\mid \frac{\overline{\alpha }{\mathrm{z}}_1+\alpha {\overline{\mathrm{z}}}_1+2\mathrm{c}}{2\mid \alpha \mid }\mid$$

Answered by ajfour last updated on 13/Aug/17

$${\mathrm{z}}_{\mathrm{F}}=\alpha (\mathrm{see}\mathrm{Q}.19592)$$$${\mathrm{z}}_0={\mathrm{z}}_1-\lambda \alpha \mathrm{and}{\overline{\mathrm{z}}}_0={\overline{\mathrm{z}}}_1-\lambda \overline{\alpha }$$$$\mathrm{as}{\mathrm{z}}_0\mathrm{lies}\mathrm{on}\mathrm{line}\overline{\alpha }\mathrm{z}+\alpha \overline{\mathrm{z}}+2\mathrm{c}=0$$$$\mathrm{we}\mathrm{have}\overline{\alpha }({\mathrm{z}}_1-\lambda \alpha )+\alpha ({\overline{\mathrm{z}}}_1-\lambda \overline{\alpha })+2\mathrm{c}=0$$$$\mathrm{or}\overline{\alpha }{\mathrm{z}}_1+\alpha {\overline{\mathrm{z}}}_1+2\mathrm{c}=2\lambda \mid \alpha {\mid }^2...\left(\mathrm{i}\right)$$$$\mathrm{p}=\mid {\mathrm{z}}_1-{\mathrm{z}}_0\mid =\mid \lambda \alpha \mid =\mid \lambda \mid \mid \alpha \mid$$$$\mathrm{using}\left(\mathrm{i}\right)\mathrm{we}\mathrm{get}:$$$$\mathrm{p}=\mid {\mathrm{z}}_1-{\mathrm{z}}_0\mid =\mid \frac{\overline{\alpha }{\mathrm{z}}_1+\alpha {\overline{\mathrm{z}}}_1+2\mathrm{c}}{2\mid \alpha \mid }\mid .$$

Commented by Tinkutara last updated on 13/Aug/17

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

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