(1+i)²=2i
(1+i)^4=(2i)²=-4
(1+i)^8=16
(1+i)^12=-64
presupunem ca (1+i)^4k= (-1)^k *4^k
care se verificvca pt k=1
presu[punem adeavarat pt k
verifixcam pt k->k=1
(1+i)^ (4k+4)=(1+i)^4k * (1+i) ^4= (-1)^k *4^k*(-4)= (-1)^ (k+1) * 4^ (k+1)
Pk->Pk+1, formula este verificat pruin inductie
deci
(1+i)^4k= (-1)^k *4^k
adica , pt n=4k
z^n=(1+i)^n= (-1)^n/4* 4^k
si atunci
(z^n)²=1*(4^k)²=4^2k =(2²)^2k=2^4k=2^n
