The modulus of the complex number (1+i)^2(1+3 i) (2-6 i)(2-2 i) is

Step-by-step solution

Here, (1+i)^2(1+3 i) (2-6 i)(2-2 i) aligned & = (1+i^2+2 i )(1+3 i) 4-4 i-12 i+12 i^2 & = 2 i+6 i^2 -8-16 i = 2 i-6 -8-16 i & = i-3 -4-8 i & = 3-i (4+8 i) (4-8 i) (4-8 i) & = 12-28 i-8 16+64 & = 4-28 i 80 = 1-7 i 20 & = 1 20 - 7 i 20 aligned Let z=x+i y |z|= x^2+y^2 aligned & |z|= ( 1 20 )^2+ ( 7 20 )^2 & |z|= 1 20 50 = 5 20 2 & |z|= 2 4 aligned