If-Re-z-4-2z-1-1-2-then-z-is-represented-by-a-point-lying-on-1-A-circle-2-An-ellipse-3-A-straight-line-4-No-real-locus-

Question Number 20800 by Tinkutara last updated on 03/Sep/17

$$\mathrm{If}\:\mathrm{Re}\left(\frac{{z}\:+\:\mathrm{4}}{\mathrm{2}{z}\:−\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{then}\:{z}\:\mathrm{is}\:\mathrm{represented} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{point}\:\mathrm{lying}\:\mathrm{on} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{A}\:\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{An}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{A}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{real}\:\mathrm{locus} \\ $$

Commented by ajfour last updated on 03/Sep/17

$${Re}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$\Rightarrow\:{straight}\:{line}. \\ $$

Answered by ajfour last updated on 03/Sep/17

⇒ ((x+4+iy)/(2x−1+i2y))=(1/2)+iq ⇒ (((x+4+iy)(2x−1−i2y))/((2x−1)^2 +4y^2 ))=(1/2)+iq ⇒ (((x+4)(2x−1)+2y^2 )/((2x−1)^2 +4y^2 )) =(1/2) or ((2(x+4)(2x−1)+4y^2 )/((2x−1)^2 +4y^2 )) = 1 ⇒ 2x−1=0 or 2(x+4)=2x−1 from first x= Re(z)=(1/2) from second 2x+8=2x−1 no solution. so Re(z)=(1/2) Im(z)≠0 . straight line with one point (z=1/2+0i) removed.

$$\Rightarrow\:\:\:\:\:\:\:\frac{{x}+\mathrm{4}+{iy}}{\mathrm{2}{x}−\mathrm{1}+{i}\mathrm{2}{y}}=\frac{\mathrm{1}}{\mathrm{2}}+{iq} \\ $$$$\Rightarrow\:\frac{\left({x}+\mathrm{4}+{iy}\right)\left(\mathrm{2}{x}−\mathrm{1}−{i}\mathrm{2}{y}\right)}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}+{iq} \\ $$$$\Rightarrow\:\:\frac{\left({x}+\mathrm{4}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{2}{y}^{\mathrm{2}} }{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\:\:\:\frac{\mathrm{2}\left({x}+\mathrm{4}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)+\mathrm{4}{y}^{\mathrm{2}} }{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\:\:\:{or}\:\:\mathrm{2}\left({x}+\mathrm{4}\right)=\mathrm{2}{x}−\mathrm{1} \\ $$$${from}\:{first}\:\:\:\boldsymbol{{x}}=\:\boldsymbol{{Re}}\left(\boldsymbol{{z}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${from}\:{second}\:\:\mathrm{2}{x}+\mathrm{8}=\mathrm{2}{x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{no}\:{solution}. \\ $$$$\:\:\:\:\:{so}\:\:{Re}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:{Im}\left({z}\right)\neq\mathrm{0}\:. \\ $$$${straight}\:{line}\:{with}\:{one}\:{point} \\ $$$$\left({z}=\mathrm{1}/\mathrm{2}+\mathrm{0}{i}\right)\:\:\:{removed}. \\ $$

Commented by ajfour last updated on 03/Sep/17