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Question-197011

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Question Number 197011 by sonukgindia last updated on 06/Sep/23
Answered by MM42 last updated on 06/Sep/23
let u=m+ni & v=k+li (e^(−(a/2)) −i)(m+ni)+(e^(−(a/2)) +i)(k+li) ⇒e^(−(a/2)) m+n+e^(−(a/2)) k−l=(m+k)e^(−(a/2)) +n−l=2(√3) (n+l)e^(−(a/2)) −m+k=0 m+k=0 & n−l=2(√3) n+l=0 & k−m=0 m=k=0 & n=(√3) & l=−(√3) ⇒u=(√3)i & v=−(√3)i
$${let}\:\:{u}={m}+{ni}\:\:\:\&\:\:\:{v}={k}+{li} \\ $$$$\left({e}^{−\frac{{a}}{\mathrm{2}}} −{i}\right)\left({m}+{ni}\right)+\left({e}^{−\frac{{a}}{\mathrm{2}}} +{i}\right)\left({k}+{li}\right) \\ $$$$\Rightarrow{e}^{−\frac{{a}}{\mathrm{2}}} {m}+{n}+{e}^{−\frac{{a}}{\mathrm{2}}} {k}−{l}=\left({m}+{k}\right){e}^{−\frac{{a}}{\mathrm{2}}} +{n}−{l}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left({n}+{l}\right){e}^{−\frac{{a}}{\mathrm{2}}} −{m}+{k}=\mathrm{0} \\ $$$${m}+{k}=\mathrm{0}\:\:\&\:\:\:{n}−{l}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${n}+{l}=\mathrm{0}\:\:\:\&\:\:{k}−{m}=\mathrm{0} \\ $$$${m}={k}=\mathrm{0}\:\:\&\:\:\:{n}=\sqrt{\mathrm{3}}\:\&\:{l}=−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{u}=\sqrt{\mathrm{3}}{i}\:\:\:\:\&\:\:\:{v}=−\sqrt{\mathrm{3}}{i} \\ $$$$ \\ $$
Answered by Frix last updated on 06/Sep/23
Let e^(−(a/2)) =k ⇒ v=((2(√3)−(k−i)u)/(k+i))=((2(√3)k−(k^2 −1)u)/(k^2 +1))+((2(ku−(√3)))/(k^2 +1))i Let u=x+yi; x, y ∈R v=((2k(y−(√3))−(k^2 −1)x)/(k^2 +1))+((2kx−(k^2 −1)y−2(√3))/(k^2 +1))i ⇒ u=x+yi v=(((e^a −1)x−2e^(a/2) (y−(√3)))/(e^a +1))+((2e^(a/2) x+(e^a −1)y−2(√3)e^a )/(e^a +1))i
$$\mathrm{Let}\:\mathrm{e}^{−\frac{{a}}{\mathrm{2}}} ={k}\:\Rightarrow \\ $$$${v}=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\left({k}−\mathrm{i}\right){u}}{{k}+\mathrm{i}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{k}−\left({k}^{\mathrm{2}} −\mathrm{1}\right){u}}{{k}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}\left({ku}−\sqrt{\mathrm{3}}\right)}{{k}^{\mathrm{2}} +\mathrm{1}}\mathrm{i} \\ $$$$\mathrm{Let}\:{u}={x}+{y}\mathrm{i};\:{x},\:{y}\:\in\mathbb{R} \\ $$$${v}=\frac{\mathrm{2}{k}\left({y}−\sqrt{\mathrm{3}}\right)−\left({k}^{\mathrm{2}} −\mathrm{1}\right){x}}{{k}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}{kx}−\left({k}^{\mathrm{2}} −\mathrm{1}\right){y}−\mathrm{2}\sqrt{\mathrm{3}}}{{k}^{\mathrm{2}} +\mathrm{1}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${u}={x}+{y}\mathrm{i} \\ $$$${v}=\frac{\left(\mathrm{e}^{{a}} −\mathrm{1}\right){x}−\mathrm{2e}^{\frac{{a}}{\mathrm{2}}} \left({y}−\sqrt{\mathrm{3}}\right)}{\mathrm{e}^{{a}} +\mathrm{1}}+\frac{\mathrm{2e}^{\frac{{a}}{\mathrm{2}}} {x}+\left(\mathrm{e}^{{a}} −\mathrm{1}\right){y}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{e}^{{a}} }{\mathrm{e}^{{a}} +\mathrm{1}}\mathrm{i} \\ $$

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