Question

2(1-3i)la puterea 2 determinati modulul numarului complex z

Answer 1

2(1-6i+9i^2)
i^2=-1
2(1-6i-9)=2(-8-6i)=-16-12i 
modul de z=√256+144=√400=20

Answer 2

$2\cdot(1-3i)^2 = 2\cdot(1^2-2\cdot 1\cdot 3i+9i^2) = 2\cdot (1-6i-9) = \\ =2\cdot(-8-6i) \\ \\ \big|2\cdot(-8-6i)\big| = |2|\cdot|-8-6i| = 2\cdot |-8-6i| = 2\cdot |8+6i| = \\ =2\cdot \sqrt{8^2+6^2} = 2\cdot \sqrt{64+36} = 2\cdot \sqrt{100} = 2\cdot 10 = 20$