Question

Sa se rezolve ecuatiile:
1) z¹⁰-(1+21)z⁵+i+1= 0
2)1+ (z+i/z-i) + (z+i/z-i)² + (z+1/z-1)³=0
3)1+ z+z²+ ... +zⁿ=0
Am nevoie de ajutor!!!! Mersii mult!!!

Answer 1

3.) inmultesti  ambii  membrii  ai ec.cu (z-1) Dupa  efectuarea inmultirilor  si a  reducerilor  vei  obtine
z^(n+1)-1=0
z^(n+1)=0 Radacina de  ordin n+1 a  unitatii.
Aplici formula
Zk=cos(2kπ/(n+1)+isin2kπ/(n+1  k∈[0 1,2...n]
zo=1
z1=cos2π/(n+1)+isin2π/(n+1)
z2=cos4π+isin4π
.....................................
Zn=cos2nπ/(n+1)+isin2nπ/(n+1)
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Notezi  t=(z+i)/(z-i)   t∈C
1+t+t²+t³=0
(1+t)*(1+t²)=0=>
1+t=0=>
1+(z+i)/(z-i)=0
z-i+z+i=0  2z=0  z=0
1+t²=0  t²=-1   t=+/-i
(z+1)/(z-1)=-i
z+1=t1=-iz+i
z+iz=i+1
(1+i)*z=i+1 => z1=1
z2=t2=i
(z+i)(z-i)=i
z+i=z*i-i²
z+i=zi+1
z-zi=1-i
z(1-i)=1-i
z2=1