Question Number 19739 by Tinkutara last updated on 15/Aug/17
$$\mathrm{If}\:{z}\:=\:{x}\:+\:{iy}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number} \\ $$$$\mathrm{satisfying}\:\mid{z}\:+\:\frac{{i}}{\mathrm{2}}\mid^{\mathrm{2}} \:=\:\mid{z}\:−\:\frac{{i}}{\mathrm{2}}\mid^{\mathrm{2}} ,\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 15/Aug/17
$$\:\:\:\:\:\:\:\:\:\:\mathrm{y}=\mathrm{0}\:\:\:\:\left(\mathrm{real}\:\mathrm{axis}\right) \\ $$$$\left(\mathrm{z}+\frac{\mathrm{i}}{\mathrm{2}}\right)\left(\bar {\mathrm{z}}−\frac{\mathrm{i}}{\mathrm{2}}\right)=\left(\mathrm{z}−\frac{\mathrm{i}}{\mathrm{2}}\right)\left(\bar {\mathrm{z}}+\frac{\mathrm{i}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\mathrm{z}\bar {\mathrm{z}}−\frac{\mathrm{iz}}{\mathrm{2}}+\frac{\mathrm{i}\bar {\mathrm{z}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{z}\bar {\mathrm{z}}+\frac{\mathrm{iz}}{\mathrm{2}}−\frac{\mathrm{i}\bar {\mathrm{z}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{z}−\bar {\mathrm{z}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{y}=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 15/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$