Let \(u,v\) be distinct complex numbers. If \(u^2=v\) and \(v^2=u,\) what is \(uv?\)
I got 1
\(\text{Let u, v be distinct complex numbers. }\\ If\;\;u^2=v\;\;and\;\;v^2=u, \;\; what \;\;is \;\;uv\\ \)
\(uv=Ae^{\alpha i }*Be^{\beta i}\\ uv=ABe^{(\alpha +\beta) i}\\\)
Let
\(u=Ae^{\alpha i}\qquad \qquad v=Be^{\beta i}\\ u^2=A^2e^{2\alpha i}\qquad \quad v^2=B^2e^{2\beta i}\\~\\ u=v^2 \qquad \qquad \qquad v=u^2 \\ Ae^{\alpha i}=B^2e^{2\beta i}\quad \quad Be^{\beta i}=A^2e^{2\alpha i}\\~\\ Ae^{\alpha i}Be^{\beta i}=B^2e^{2\beta i}(A^2e^{2\alpha i})\\ ABe^{\alpha i+\beta i}=A^2B^2e^{2\alpha i+2\beta i}\\ e^{\alpha i+\beta i - ( 2\alpha i+2\beta i) }=AB\\ AB=e^{-i(\alpha +\beta ) }\\ \)
so
\(uv=ABe^{(\alpha +\beta) i}\\ uv=e^{-(\alpha+\beta)i}*e^{(\alpha +\beta) i}\\ uv=e^0\\ uv=1\)
Coding:
\text{Let u, v be distinct complex numbers. }\\
If\;\;u^2=v\;\;and\;\;v^2=u, \;\; what \;\;is \;\;uv\\
uv=Ae^{\alpha i }*Be^{\beta i}\\
u=Ae^{\alpha i}\qquad \qquad v=Be^{\beta i}\\
u^2=A^2e^{2\alpha i}\qquad \quad v^2=B^2e^{2\beta i}\\~\\
u=v^2 \qquad \qquad \qquad v=u^2 \\
Ae^{\alpha i}=B^2e^{2\beta i}\quad \quad Be^{\beta i}=A^2e^{2\alpha i}\\~\\
Ae^{\alpha i}Be^{\beta i}=B^2e^{2\beta i}(A^2e^{2\alpha i})\\
ABe^{\alpha i+\beta i}=A^2B^2e^{2\alpha i+2\beta i}\\
e^{\alpha i+\beta i - ( 2\alpha i+2\beta i) }=AB\\
AB=e^{-i(\alpha +\beta ) }\\
uv=ABe^{(\alpha +\beta) i}\\
uv=e^{-(\alpha+\beta)i}*e^{(\alpha +\beta) i}\\
uv=e^0\\
uv=1
u^2 = v
v^2 = u subtract these
u^2 - v^2 = v - u
( u - v) (u + v) = v - u
(u - v) ( u + v) = - ( u -v) divide both sides by (u - v)
(u + v) = -1
(v + u) = -1
square both sides
u^2 + 2uv + v^2 = 1
(u^2 + v^2) + 2uv = 1
Adding the first two equations and we get that u^2 + v^2 = v + u....so....
(v + u) + 2uv = 1
-1 + 2uv = 1 add 1 to both sides
2uv = 2
uv = 1