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Question Number 19688 by Tinkutara last updated on 14/Aug/17

$$\mathrm{The}\mathrm{vertices}\mathrm{of}\mathrm{a}\mathrm{square}\mathrm{are}{z}_1,z_2,z_3$$$$\mathrm{and}{z}_4\mathrm{taken}\mathrm{in}\mathrm{the}\mathrm{anticlockwise}\mathrm{order},$$$$\mathrm{then}{z}_3=$$$$\left(1\right)-{iz}_1+(1+i)z_2$$$$\left(2\right){iz}_1+(1+i)z_2$$$$\left(3\right)z_1+(1+i)z_2$$$$\left(4\right)(1+i)z_1+z_2$$

Answered by ajfour last updated on 14/Aug/17

$${\mathrm{z}}_3={\mathrm{z}}_1+({\mathrm{z}}_2-{\mathrm{z}}_1){\mathrm{e}}^{\mathrm{i}\pi /4}\times \sqrt{2}$$$$={\mathrm{z}}_1+({\mathrm{z}}_2-{\mathrm{z}}_1)(1+\mathrm{i})$$$${\mathrm{z}}_3=-{\mathrm{iz}}_1+(1+\mathrm{i}){\mathrm{z}}_2.\left(1\right)$$

Commented by Tinkutara last updated on 15/Aug/17

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

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