|z| = 4
==> |(1+i)^3(a+bi)/(1-i)| = 4
==> |1+i|³ * |a+bi|/|1-i| =4
==> √(a²+b²) =2 ==> a²+b² =4
|w+2-2i|
= |a+bi +2 -2i|
= √[(a+2)²+(b-2)²]
= √[(a²+b²)+4(a-b)+8]
= √[12+4(a-b)]
又 (a-b)² = a²+b² -2ab ≤2(a²+b² ) =8
==> |a-b|≤2√2
==> -2√2 ≤ a-b ≤ 2√2
∴ √(12-8√2) ≤ |w+2-2i| = √[12+4(a-b)] ≤√(12+8√2)
==> 2√2 - 2 ≤ |w+2-2i| ≤ 2√2 +2