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Question Number 47543 by maxmathsup by imad last updated on 11/Nov/18
solve (1+ix)^n =n   with x unknown real and n integr natural .
solve(1+ix)n=nwithxunknownrealandnintegrnatural.
Answered by arcana last updated on 11/Nov/18
ln(e^((1+ix)n) )=n ln(e e^(ix) )=n  ln(e)+ln(e^(ix) )=1+ln(e^(ix) )=1  ln(e^(ix) )=0=ln(1)⇒e^(ix) =cos(x)+isin(x)=1+i0  si y solo si cos(x)=1, sin(x)=0  x=kπ con k∈Z  verificar deduccion
ln(e(1+ix)n)=nln(eeix)=nln(e)+ln(eix)=1+ln(eix)=1ln(eix)=0=ln(1)eix=cos(x)+isin(x)=1+i0siysolosicos(x)=1,sin(x)=0x=kπconkZverificardeduccion
Commented by Smail last updated on 11/Nov/18
(1+ix)^n =ln(e^((1+ix)^n ) ) not ln(e^((1+ix)n) )
(1+ix)n=ln(e(1+ix)n)notln(e(1+ix)n)
Answered by Smail last updated on 11/Nov/18
(1+ix)^n =ne^(2ikπ)   with k=(0,1,2,...,n−1)  1+ix=(n)^(1/n) e^((2ikπ)/n)   1+ix=(n)^(1/n) cos(((2kπ)/n))+i(n)^(1/n) sin(((2kπ)/n))  (n)^(1/n) cos(((2kπ)/n))=1 and  (n)^(1/n) sin(((2kπ)/n))=x  ((2kπ)/n)=cos^(−1) ((1/( (n)^(1/n) ))) +2mπ with  m∈IN  x=(n)^(1/n) sin(cos^(−1) ((1/( (n)^(1/n) )))+2mπ)  =(n)^(1/n) (√(1−(1/( (n^2 )^(1/n) ))))=(√((n^2 )^(1/n) −1))  x=(√((n^2 )^(1/n) −1))
(1+ix)n=ne2ikπwithk=(0,1,2,,n1)1+ix=nne2ikπn1+ix=nncos(2kπn)+innsin(2kπn)nncos(2kπn)=1andnnsin(2kπn)=x2kπn=cos1(1nn)+2mπwithmINx=nnsin(cos1(1nn)+2mπ)=nn11n2n=n2n1x=n2n1

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