Show that the modulus of (1-i)^6/(1+i)=4√2?

To simplify the expression, we first need to find the value of (1-i)^6 and (1+i).

(1-i)^6:
(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i
(1-i)^4 = (-2i)^2 = 4
(1-i)^6 = (1-i)^4 * (1-i)^2 = 4 * -2i = -8i

(1+i):
(1+i) = 1 + i

Now, we can substitute these values back into the original expression:

(1-i)^6 / (1+i) = (-8i) / (1+i) = -8i / (1+i) * (1-i) / (1-i) = -8i(1-i) / (1^2 - i^2) = -8i(1-i) / (1 + 1) = -4(1-i) = -4 + 4i

To find the modulus of this complex number, we use the formula:

|a + bi| = √(a^2 + b^2)

So, the modulus of -4 + 4i is:

|-4 + 4i| = √((-4)^2 + 4^2) = √(16 + 16) = √32 = 4√2

Therefore, the modulus of (1-i)^6 / (1+i) is 4√2.