$z= (i-1) + \bigg [ 1+(1+i) +(1+i)^2 + ... + (1+i)^{2011} \bigg ] = (i-1) +A$
Tính $A$ theo cấp số nhân
$A=\dfrac{u_1 (q^n-1)}{q-1}$ với $u_1=1;\ q=(1+i)$
Vậy $A=\dfrac{(1+i)^{2011}-1}{1+i-1} = \dfrac{(1+i)(1+i)^{2010}}{i} -\dfrac{1}{i}$
$=\dfrac{(1+i)[ (1+i)^2]^{1005}}{i}+i =\dfrac{(1+i) (2i)^{1005}}{i}+i=2^{1005} + 2i +2^{1005} i =2^{1005} +(2^{1005}+1)i$
Vậy $z= i-1 +2^{1005} +(2^{1005}+1)i=(2^{1005} -1) + (2^{1005}+2)i$