... = |z |2 + z · z + z · z + |z |2 Because z · z = z · z = z · z it follows that z z + z · z = Re (z · z ) ≤ 2 |z · z | = 2 |z | · |z |, hence |z + z |2 ≤ ( |z | + |z |)2 , and consequently, |z + z ... complex numbers z = (1, 2), z = (−2, 3) and z = (1, −1) Compute the following complex numbers: a) z + z + z ; b) z z + z z + z z ; c) z z z ; 2 d) z + z + z ; e) z1 z2 z3 + + ; z2 z3 z1 f) 2 z1 + z2 ... + z ) z1 · z2 · · · zn is real 33 Let z , z , z be distinct complex numbers such that |z | = |z | = |z | > If z + z z , z + z z and z + z z are real numbers, prove that z z z = 34 Let x1 and x2...