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Question Number 113262 by bemath last updated on 12/Sep/20
find the complex form of equation 4x^2 −2y^2 =5
$$\mathrm{find}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{form}\:\mathrm{of}\: \\ $$$$\mathrm{equation}\:\mathrm{4x}^{\mathrm{2}} −\mathrm{2y}^{\mathrm{2}} =\mathrm{5} \\ $$
Answered by mr W last updated on 12/Sep/20
(x^2 /((((√5)/2))^2 ))−(y^2 /((((√5)/( (√2))))^2 ))=1 c^2 =a^2 +b^2 =(5/4)+(5/2)=((15)/4) c=((√(15))/2) it′s a hyperbola with foci (−((√(15))/2),0) and (((√(15))/2),0) as well as semi major axis a=((√5)/2). according to the definition of hyperbola we can write its equation also in complex form: ∣z+c∣−∣z−c∣=±2a or ∣z+((√(15))/2)∣−∣z−((√(15))/2)∣=±(√5)
$$\frac{{x}^{\mathrm{2}} }{\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{\left(\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$${c}=\frac{\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$${it}'{s}\:{a}\:{hyperbola}\:{with}\:{foci}\:\left(−\frac{\sqrt{\mathrm{15}}}{\mathrm{2}},\mathrm{0}\right) \\ $$$${and}\:\left(\frac{\sqrt{\mathrm{15}}}{\mathrm{2}},\mathrm{0}\right)\:{as}\:{well}\:{as}\:{semi}\:{major} \\ $$$${axis}\:{a}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}. \\ $$$${according}\:{to}\:{the}\:{definition}\:{of} \\ $$$${hyperbola}\:{we}\:{can}\:{write}\:{its}\:{equation} \\ $$$${also}\:{in}\:{complex}\:{form}: \\ $$$$\mid{z}+{c}\mid−\mid{z}−{c}\mid=\pm\mathrm{2}{a} \\ $$$${or} \\ $$$$\mid{z}+\frac{\sqrt{\mathrm{15}}}{\mathrm{2}}\mid−\mid{z}−\frac{\sqrt{\mathrm{15}}}{\mathrm{2}}\mid=\pm\sqrt{\mathrm{5}} \\ $$
Commented by mr W last updated on 12/Sep/20
certainly you can also write the equation in many other complex forms.
$${certainly}\:{you}\:{can}\:{also}\:{write}\:{the} \\ $$$${equation}\:{in}\:{many}\:{other}\:{complex}\:{forms}. \\ $$
Commented by bemath last updated on 12/Sep/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by john santu last updated on 12/Sep/20
let z = x+iy and z^− = x−iy { ((z+z^− = (2x,0); x = ((z+z^− )/2))),((z−z^− = (0, 2y); y=((z−z^− )/2))) :} ⇔ 4x^2 −2y^2 =5 in complex form 4(((z+z^− )/2))^2 −2(((z−z^− )/2))^2 = 5 (z+z^− )^2 −(((z−z^− )^2 )/2) = 5 2(z+z^− )^2 −(z−z^− )^2 = 10
$${let}\:{z}\:=\:{x}+{iy}\:{and}\:\overset{−} {{z}}\:=\:{x}−{iy} \\ $$$$\:\begin{cases}{{z}+\overset{−} {{z}}\:=\:\left(\mathrm{2}{x},\mathrm{0}\right);\:{x}\:=\:\frac{{z}+\overset{−} {{z}}}{\mathrm{2}}}\\{{z}−\overset{−} {{z}}\:=\:\left(\mathrm{0},\:\mathrm{2}{y}\right);\:{y}=\frac{{z}−\overset{−} {{z}}}{\mathrm{2}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} =\mathrm{5}\:{in}\:{complex}\:{form} \\ $$$$\:\mathrm{4}\left(\frac{{z}+\overset{−} {{z}}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{z}−\overset{−} {{z}}}{\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{5} \\ $$$$\left({z}+\overset{−} {{z}}\right)^{\mathrm{2}} −\frac{\left({z}−\overset{−} {{z}}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\:\mathrm{5} \\ $$$$\:\mathrm{2}\left({z}+\overset{−} {{z}}\right)^{\mathrm{2}} −\left({z}−\overset{−} {{z}}\right)^{\mathrm{2}} \:=\:\mathrm{10} \\ $$

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