F.4 Number System question

1.
(1+i)z2-(2+3i)z+2+i=0
[ (1+i)z + 1 ][ z + (2+i)] = 0
z = -1/ 1+i or z = -(2+i)
z = -(1-i)/2 or z = -(2+i)


2.
(p+qi)2 = 21-20i
p^2 + 2pqi - q^2 = 21 - 20i

p^2 - q^2 = 21
p^2 = 21 +q^2

2pq = -20
pq = -10
p^2q^2 = 100
(21+q^2)(q^2)=100
q^4+21q^2-100=0
(q^2+25)(q^2-4)= 0
q^2 = -25(reject, p is real number) or q^2 = 4
(q = 2 and p = -5) or (q = -2 and p = 5)
since p>q
p = 5 and q = -2

square roots of 21-20i :
√ (5-2i) and √ (5-2i)