Menu Close

find-x-y-in-R-x-yi-3-1-15-i-5-3-i-

Tinku Tara 0 Comments
Pin




Question Number 59588 by aliesam last updated on 12/May/19
find x,y in R (x+yi)^3 =((1+(√(15)) i)/( (√5) − (√3) i))
$${find}\:{x},{y}\:{in}\:{R} \\ $$$$\left({x}+{yi}\right)^{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{15}}\:{i}}{\:\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{3}}\:{i}} \\ $$
Commented by maxmathsup by imad last updated on 12/May/19
⇒x+iy =^3 (√((1+(√(15))i)/( (√5)−(√3)i))) let solve Z^3 =((1+(√(15))i)/( (√5)−(√3)i)) we have ∣1+(√(15))i∣ =(√(1+15))=4 and ∣(√5)−(√3)i∣ =(√(5 +3))=2(√(2 ))⇒∣((1+(√(15))i)/( (√5)−(√3)i))∣=(4/(2(√2))) =(√(2 )) 1+(√(15))i =4{(1/4)+((√(15))/4)i} =r e^(iθ) ⇒r=4 and cosθ=(1/4) and sinθ =((√(15))/4) ⇒ tanθ =(√(15)) ⇒θ =arctan((√(15))) also (√5)−(√3)i =2(√2){((√5)/(2(√2))) −((√3)/(2(√2)))i}=r^′ e^(iθ′) ⇒ r^′ =2(√2) and θ^′ =arctan(((−(√3))/( (√5)))) ⇒arg(((1+(√(15))i)/( (√5)−(√3)i)))≡ arctan((√(15))) +arctan(((√3)/( (√5))))[2π] ⇒ ⇒((1+(√(15))i)/( (√5)−(√3)i)) =(√2) e^(i(arctan((√(15)))+arctan(((√3)/( (√5)))))) =Z^3 ⇒ Z =^6 (√2)e^((1/3)(arctan((√(15)))+arctan(((√3)/( (√5)))))i) =^6 (√2){cos((1/3)(arctan((√(15)))+arctan(((√3)/( (√5))))))+i sin((1/3)(arctan((√(15)))+arctan(((√3)/( (√5))))) ⇒ x =^6 (√2)cos((1/3)(arctan((√(15)))+arctan(((√3)/( (√5)))))) and y =^6 (√2)sin((1/3)(arctan((√(15))) +arctan(((√3)/( (√5)))))) .
$$\Rightarrow{x}+{iy}\:=^{\mathrm{3}} \sqrt{\frac{\mathrm{1}+\sqrt{\mathrm{15}}{i}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}}}\:\:\:{let}\:{solve}\:{Z}^{\mathrm{3}} \:=\frac{\mathrm{1}+\sqrt{\mathrm{15}}{i}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}}\:\:{we}\:{have}\: \\ $$$$\mid\mathrm{1}+\sqrt{\mathrm{15}}{i}\mid\:=\sqrt{\mathrm{1}+\mathrm{15}}=\mathrm{4}\:\:\:{and}\:\:\:\mid\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}\mid\:=\sqrt{\mathrm{5}\:+\mathrm{3}}=\mathrm{2}\sqrt{\mathrm{2}\:}\Rightarrow\mid\frac{\mathrm{1}+\sqrt{\mathrm{15}}{i}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}}\mid=\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}\:} \\ $$$$\mathrm{1}+\sqrt{\mathrm{15}}{i}\:=\mathrm{4}\left\{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}{i}\right\}\:={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\mathrm{4}\:{and}\:{cos}\theta=\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{sin}\theta\:=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:\Rightarrow \\ $$$${tan}\theta\:=\sqrt{\mathrm{15}}\:\Rightarrow\theta\:={arctan}\left(\sqrt{\mathrm{15}}\right)\:{also}\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}\:=\mathrm{2}\sqrt{\mathrm{2}}\left\{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}{i}\right\}={r}^{'} \:{e}^{{i}\theta'} \:\Rightarrow \\ $$$${r}^{'} \:=\mathrm{2}\sqrt{\mathrm{2}}\:{and}\:\:\theta^{'} \:={arctan}\left(\frac{−\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\:\Rightarrow{arg}\left(\frac{\mathrm{1}+\sqrt{\mathrm{15}}{i}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}}\right)\equiv\:{arctan}\left(\sqrt{\mathrm{15}}\right)\:+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\left[\mathrm{2}\pi\right]\:\Rightarrow \\ $$$$\:\Rightarrow\frac{\mathrm{1}+\sqrt{\mathrm{15}}{i}}{\:\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}{i}}\:=\sqrt{\mathrm{2}}\:{e}^{{i}\left({arctan}\left(\sqrt{\mathrm{15}}\right)+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right)} \:={Z}^{\mathrm{3}} \:\Rightarrow \\ $$$${Z}\:=^{\mathrm{6}} \sqrt{\mathrm{2}}{e}^{\frac{\mathrm{1}}{\mathrm{3}}\left({arctan}\left(\sqrt{\mathrm{15}}\right)+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right){i}} \\ $$$$=^{\mathrm{6}} \sqrt{\mathrm{2}}\left\{{cos}\left(\frac{\mathrm{1}}{\mathrm{3}}\left({arctan}\left(\sqrt{\mathrm{15}}\right)+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right)\right)+{i}\:\:{sin}\left(\frac{\mathrm{1}}{\mathrm{3}}\left({arctan}\left(\sqrt{\mathrm{15}}\right)+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right)\:\Rightarrow\right.\right. \\ $$$${x}\:=^{\mathrm{6}} \sqrt{\mathrm{2}}{cos}\left(\frac{\mathrm{1}}{\mathrm{3}}\left({arctan}\left(\sqrt{\mathrm{15}}\right)+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right)\right)\:{and} \\ $$$${y}\:=^{\mathrm{6}} \sqrt{\mathrm{2}}{sin}\left(\frac{\mathrm{1}}{\mathrm{3}}\left({arctan}\left(\sqrt{\mathrm{15}}\right)\:+{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}}}\right)\right)\right)\:. \\ $$
Answered by tanmay last updated on 12/May/19
(((1+(√(15)) i)((√5) +(√3) i))/8) (((2+2(√(15)) i)((√5) +(√3) i))/(16)) (((5−3+2(√(15)) i)((√5) +(√3) i))/(16)) ((((√5) +(√3) i)^2 ((√5) +(√3) i))/(16)) ((((√5) +(√3) i)^3 )/((2^(4/3) )^3 )) so x=((√5)/2^(4/3) ) and y=((√3)/2^(4/3) )
$$\frac{\left(\mathrm{1}+\sqrt{\mathrm{15}}\:{i}\right)\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)}{\mathrm{8}} \\ $$$$\frac{\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{15}}\:{i}\right)\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)}{\mathrm{16}} \\ $$$$\frac{\left(\mathrm{5}−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{15}}\:{i}\right)\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)}{\mathrm{16}} \\ $$$$\frac{\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)^{\mathrm{2}} \left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)}{\mathrm{16}} \\ $$$$\frac{\left(\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{3}}\:{i}\right)^{\mathrm{3}} }{\left(\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} \right)^{\mathrm{3}} } \\ $$$${so}\:{x}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} }\:\:\:{and}\:{y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}^{\frac{\mathrm{4}}{\mathrm{3}}} } \\ $$
Commented by MJS last updated on 12/May/19
your solution is z_1 =(((√5)(4)^(1/3) )/4)+(((√3)(4)^(1/3) )/4)i but there are 2 more: z_2 =z_1 (−(1/2)−((√3)/2)i)=((4)^(1/3) /8)(3−(√5))−(((√3)(4)^(1/3) )/8)(1+(√5))i and z_3 =z_1 (−(1/2)+((√3)/2)i)=−((4)^(1/3) /8)(3+(√5))−(((√3)(4)^(1/3) )/8)(1−(√5))i
$$\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:{z}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{4}}+\frac{\sqrt{\mathrm{3}}\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{4}}\mathrm{i}\:\mathrm{but}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{2}\:\mathrm{more}:\:{z}_{\mathrm{2}} ={z}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)=\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{8}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)−\frac{\sqrt{\mathrm{3}}\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{8}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{i} \\ $$$$\mathrm{and}\:{z}_{\mathrm{3}} ={z}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)=−\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{8}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)−\frac{\sqrt{\mathrm{3}}\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\mathrm{8}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\mathrm{i} \\ $$
Commented by tanmay last updated on 12/May/19
thank yousir...
$${thank}\:{yousir}… \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *