Question Number 19623 by Tinkutara last updated on 13/Aug/17
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{if}\:\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{2}}{{z}\:−\:\mathrm{3}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 13/Aug/17
![let z=x+iy ⇒ arg[((x−2+iy)/(x−3+iy))]=(π/4) ⇒ arg[(((x−2+iy)(x−3−iy))/((x−3)^2 +y^2 ))]=(π/4) ⇒ arg[(x−2)(x−3)+y^2 −iy]=(π/4) or ((−y)/((x^2 +y^2 −5x+6)))=tan (π/4) ⇒ x^2 +y^2 −5x+y+6=0 ⇒ (x−(5/2))^2 +(y+(1/2))^2 +6−((26)/4)=0 ⇒ (x−(5/2))^2 +(y+(1/2))^2 =((1/(√2)))^2 or ∣z−(5/2)+(i/2)∣=(1/(√2)) Locus is a circle with radius r=(1/(√2)) and centre z_0 =(5/2)−(i/2) .](Q19624.png)
$$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\mathrm{x}−\mathrm{2}+\mathrm{iy}}{\mathrm{x}−\mathrm{3}+\mathrm{iy}}\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\frac{\left(\mathrm{x}−\mathrm{2}+\mathrm{iy}\right)\left(\mathrm{x}−\mathrm{3}−\mathrm{iy}\right)}{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{arg}\left[\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{3}\right)+\mathrm{y}^{\mathrm{2}} −\mathrm{iy}\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{or}\:\frac{−\mathrm{y}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}\right)}=\mathrm{tan}\:\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{5x}+\mathrm{y}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{6}−\frac{\mathrm{26}}{\mathrm{4}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\left(\mathrm{x}−\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\mid\mathrm{z}−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{i}}{\mathrm{2}}\mid=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\: \\ $$$$\:\mathrm{Locus}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{r}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\: \\ $$$$\mathrm{and}\:\mathrm{centre}\:\mathrm{z}_{\mathrm{0}} =\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{i}}{\mathrm{2}}\:. \\ $$
Commented by Tinkutara last updated on 13/Aug/17
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:\pm\:{y}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{Why}\:\pm\:{y}? \\ $$
Commented by ajfour last updated on 13/Aug/17
$$\mathrm{we}\:\mathrm{can}\:\mathrm{take}\:\mathrm{arg}\left(\frac{\mathrm{z}−\mathrm{2}}{\mathrm{z}−\mathrm{3}}\right)=\pm\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by Tinkutara last updated on 13/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 13/Aug/17
