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Q-25444-another-solution-z-1-iz-2-z-2-let-z-x-iy-x-iy-1-i-x-2-y-2-2xy-x-2-y-2-or-x-y-2-x-1-and-y-2-x-2-y-let-y-x-u-and-y-x-v-2u-2-v-u-2-i-

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Question Number 25709 by ajfour last updated on 13/Dec/17
Q. 25444 (another solution) z^� +1 =iz^2 +∣z∣^2 let z=x+iy , ⇒ x−iy+1=i(x^2 −y^2 )−2xy+x^2 +y^2 or (x−y)^2 =x+1 ,and (y^2 −x^2 )=y let y−x=u and y+x=v ⇒ 2u^2 =v−u+2 ...(i) 2uv=u+v ....(ii) ⇒ v=(u/(2u−1)) , substituting in (i) 2u^2 +u=(u/(2u−1))+2 u=1 satisfies this , so ⇒ (2u+3)(u−1)=(u/(2u−1))−1 ⇒ (2u+3)(u−1)=(((1−u))/(2u−1)) ⇒ u=1 or (2u+3)(2u−1)+1=0 or 4u^2 +4u−2=0 or 2u^2 +2u−1=0 ⇒ u=((−2±(√(4+8)))/4) =((−1±(√3))/2) for u_1 =1 , v_1 =(u_1 /(2u_1 −1))=1 so x_1 =((v_1 −u_1 )/2)=0 and y_1 =((v_1 +u_1 )/2) =1 hence z_1 =i for u_2 =(((√3)−1)/2) , v_2 =(((√3)−1)/(2((√3)−1−1))) =−((((√3)−1)(2+(√3)))/2) =−((((√3)+1))/2) x_2 =((v_2 −u_2 )/2)=−((√3)/2) y_2 =((v_2 +u_2 )/2) =−(1/2) hence z_2 =−(1/2)((√3)+i) for u_3 =−((((√3)+1))/2) v_3 =(u_3 /(2u_3 −1)) =((((√3)+1))/(2((√3)+2))) =((((√3)+1)(2−(√3)))/2) =(((√3)−1)/2) x_3 =((v_3 −u_3 )/2)=((√3)/2) and y_3 =((v_3 +u_3 )/2)=−(1/2) ⇒ z_3 =(1/2)((√3)−i) So to summarize, z_1 =i , z_2 =−(1/2)((√3)+i), and z_3 =(1/2)((√3)−i) .
$${Q}.\:\mathrm{25444}\:\:\left({another}\:{solution}\right) \\ $$$$\:\:\bar {{z}}+\mathrm{1}\:={iz}^{\mathrm{2}} +\mid{z}\mid^{\mathrm{2}} \\ $$$${let}\:\:{z}={x}+{iy}\:,\:\Rightarrow \\ $$$${x}−{iy}+\mathrm{1}={i}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${or}\:\:\left({x}−{y}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:,{and}\: \\ $$$$\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)={y} \\ $$$${let}\:\:\:{y}−{x}={u}\:\:{and}\:\:{y}+{x}={v} \\ $$$$\Rightarrow\:\:\mathrm{2}{u}^{\mathrm{2}} ={v}−{u}+\mathrm{2}\:\:\:…\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{uv}={u}+{v}\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}\:\:,\:{substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +{u}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}+\mathrm{2} \\ $$$${u}=\mathrm{1}\:{satisfies}\:{this}\:,\:{so} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}−\mathrm{1} \\ $$$$\Rightarrow\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{\left(\mathrm{1}−{u}\right)}{\mathrm{2}{u}−\mathrm{1}} \\ $$$$\Rightarrow\:\:{u}=\mathrm{1}\:{or}\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{2}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:{u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{4}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${for}\:{u}_{\mathrm{1}} =\mathrm{1}\:\:,\:{v}_{\mathrm{1}} =\frac{{u}_{\mathrm{1}} }{\mathrm{2}{u}_{\mathrm{1}} −\mathrm{1}}=\mathrm{1} \\ $$$${so}\:\:{x}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} −{u}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{0}\:\:\:{and}\: \\ $$$$\:\:\:\:\:{y}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} +{u}_{\mathrm{1}} }{\mathrm{2}}\:=\mathrm{1} \\ $$$${hence}\:\:{z}_{\mathrm{1}} ={i} \\ $$$${for}\:{u}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\:,\:\:{v}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:{x}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} −{u}_{\mathrm{2}} }{\mathrm{2}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:{y}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} +{u}_{\mathrm{2}} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${hence}\:\:\:{z}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+{i}\right) \\ $$$${for}\:\:{u}_{\mathrm{3}} =−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:{v}_{\mathrm{3}} =\frac{{u}_{\mathrm{3}} }{\mathrm{2}{u}_{\mathrm{3}} −\mathrm{1}}\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} −{u}_{\mathrm{3}} }{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and} \\ $$$$\:{y}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} +{u}_{\mathrm{3}} }{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{z}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−{i}\right) \\ $$$${So}\:\:{to}\:{summarize}, \\ $$$$\boldsymbol{{z}}_{\mathrm{1}} =\boldsymbol{{i}}\:\:,\:\:\boldsymbol{{z}}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\boldsymbol{{i}}\right),\:{and} \\ $$$$\:\boldsymbol{{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{i}}\right)\:. \\ $$
Commented by Rasheed.Sindhi last updated on 13/Dec/17
W^ elldone Sir!
$$\mathcal{W}^{\:} {elldone}\:\mathcal{S}{ir}! \\ $$

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