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Let $$u,v$$ be distinct complex numbers. If $$u^2=v$$ and $$v^2=u,$$ what is $$uv?$$

Jan 10, 2020

#1
+108771
+2

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}\\

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

Jan 10, 2020
#2
+109739
+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

Jan 10, 2020
edited by CPhill  Jan 10, 2020