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Question Number 19688 by Tinkutara last updated on 14/Aug/17
The vertices of a square are z_1 , z_2 , z_3   and z_4  taken in the anticlockwise order,  then z_3  =  (1) −iz_1  + (1 + i)z_2   (2) iz_1  + (1 + i)z_2   (3) z_1  + (1 + i)z_2   (4) (1 + i)z_1  + z_2
$$\mathrm{The}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{are}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \\ $$$$\mathrm{and}\:{z}_{\mathrm{4}} \:\mathrm{taken}\:\mathrm{in}\:\mathrm{the}\:\mathrm{anticlockwise}\:\mathrm{order}, \\ $$$$\mathrm{then}\:{z}_{\mathrm{3}} \:= \\ $$$$\left(\mathrm{1}\right)\:−{iz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{iz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{z}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \\ $$
Answered by ajfour last updated on 14/Aug/17
z_3 =z_1 +(z_2 −z_1 )e^(iπ/4) ×(√2)       =z_1 +(z_2 −z_1 )(1+i)     z_3  =−iz_1 +(1+i)z_2  .      (1)
$$\mathrm{z}_{\mathrm{3}} =\mathrm{z}_{\mathrm{1}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)\mathrm{e}^{\mathrm{i}\pi/\mathrm{4}} ×\sqrt{\mathrm{2}}\: \\ $$$$\:\:\:\:=\mathrm{z}_{\mathrm{1}} +\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\:\:\:\mathrm{z}_{\mathrm{3}} \:=−\mathrm{iz}_{\mathrm{1}} +\left(\mathrm{1}+\mathrm{i}\right)\mathrm{z}_{\mathrm{2}} \:.\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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