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Question Number 197011 by sonukgindia last updated on 06/Sep/23

Answered by MM42 last updated on 06/Sep/23

$$letu=m+ni\mathrm{\&}v=k+li$$$$(e^{-\frac{a}{2}}-i)(m+ni)+(e^{-\frac{a}{2}}+i)(k+li)$$$$\Rightarrow e^{-\frac{a}{2}}m+n+e^{-\frac{a}{2}}k-l=(m+k)e^{-\frac{a}{2}}+n-l=2\sqrt{3}$$$$(n+l)e^{-\frac{a}{2}}-m+k=0$$$$m+k=0\mathrm{\&}n-l=2\sqrt{3}$$$$n+l=0\mathrm{\&}k-m=0$$$$m=k=0\mathrm{\&}n=\sqrt{3}\mathrm{\&}l=-\sqrt{3}$$$$\Rightarrow u=\sqrt{3}i\mathrm{\&}v=-\sqrt{3}i$$$$$$

Answered by Frix last updated on 06/Sep/23

$$\mathrm{Let}{\mathrm{e}}^{-\frac{a}{2}}=k\Rightarrow$$$$v=\frac{2\sqrt{3}-(k-\mathrm{i})u}{k+\mathrm{i}}=\frac{2\sqrt{3}k-(k^2-1)u}{k^2+1}+\frac{2(ku-\sqrt{3})}{k^2+1}\mathrm{i}$$$$\mathrm{Let}u=x+y\mathrm{i};x,y\in \mathbb{R}$$$$v=\frac{2k(y-\sqrt{3})-(k^2-1)x}{k^2+1}+\frac{2kx-(k^2-1)y-2\sqrt{3}}{k^2+1}\mathrm{i}$$$$\Rightarrow$$$$u=x+y\mathrm{i}$$$$v=\frac{({\mathrm{e}}^a-1)x-{2\mathrm{e}}^{\frac{a}{2}}(y-\sqrt{3})}{{\mathrm{e}}^a+1}+\frac{{2\mathrm{e}}^{\frac{a}{2}}x+({\mathrm{e}}^a-1)y-2\sqrt{3}{\mathrm{e}}^a}{{\mathrm{e}}^a+1}\mathrm{i}$$

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