Question

Exercițiul 8 va rog cei care știți...

Answer 1

Explicație pas cu pas:

Pentru a):

Aducem z la o forma mai frumoasa:

$z=\sqrt2( {\frac{\sqrt2}{1+i} })^{103}+i^{102}=\frac{\sqrt2*\sqrt2^{103}}{(1+i)^{103}} +i^{102}=\frac{\sqrt2^{104}}{(1+i)^{103}} +i^{102}=\frac{2^{52}}{(1+i)^{103}} +i^{102}$

Vedem cat face $(1+i)^{103}$:

$(1+i)^2=1+2i+i^2=1+2i-1=2i$

Atunci:

$(1+i)^{103}=(1+i)^{102}*(1+i)=[(1+i)^2]^{51}*(1+i)=(2i)^{51}*(1+i)=2^{51}*i^{51}*(1+i)=2^{51}*(i^2)^{25}*i*(1+i)=2^{51}*(-1)^{25}*i*(1+i)=2^{51}*(-i)*(1+i)=2^{51}*(-i-i^2)=2^{51}*(1-i)$

Vedem cat face $i^{102}$:

$i^{102}=(i^2)^{51}=(-1)^{51}=-1$

Atunci, z devine:

$z=\frac{2^{52}}{(1+i)^{103}} +i^{102}=\frac{2^{52}}{2^{51}*(1-i)}-1=\frac{2}{1-i}-1=\frac{2-(1-i)}{(1-i)}  =\frac{1+i}{1-i} =\frac{(1+i)^{2}}{(1-i)*(1+i)} =\frac{2i}{1-i^2}=\frac{2i}{2}=i$

Pentru b):

Calculam intai:

$\frac{1-i}{1+i} =\frac{(1-i)^{2}}{(1-i)*(1+i)} =\frac{-2i}{1-i^2}=\frac{-2i}{2}=-i$

Atunci, z devine:

$z=1+\frac{1-i}{1+i} +(\frac{1-i}{1+i} )^2+...+(\frac{1-i}{1+i} )^9=1+(-i)+(-i)^2+(-i)^3+(-i)^4+(-i)^5+(-i)^6+(-i)^7+(-i)^8+(-i)^9=1-i+i^2-i^3+i^4-i^5+i^6-i^7+i^8-i^9=1-i-1-(i^2)*i+(i^2)^2-(i^2)^2*i+(i^2)^3-(i^2)^3*i+(i^2)^4-(i^2)^4*i=1-i-1+i+1-i-1+i+1-i=1-i$