Calculate the modulus of z = [( 2 + ( 2)^1/2)^1/2 + i*( 2 - ( 2)^1/2)^1/2]^6?

To find modulus of z =  [( 2 + ( 2)^1/2)^1/2 + i*( 2 - (
2)^1/2)^1/2]^6?.

We know cospi/4 = 2cos^2 (pi/8)
-1.

So 2cos^2(pi/8) = cospi/4+1 =
(1/2^(1/2))+1

2cos^2(pi/8) = {1+(2^(1/2)}/2^(1/2) =
{2+2^(1/2)}/2

cos^2(pi/8) =
{(2+2^(1/2)}/4.

cos(pi/8) =
(1/2)[2+2(1/2)]^(1/2)....(1)

Therefore
sin(pi /8) = {1-(1/4)[2+2^(1/2)]}^(1/2)

sinpi/8 = (1/2) {
4-2 -2^(1/2)}^(1/2)

sin(pi/8) = (1/2)
[2-2^(1/2)]^(1/2).....(2).

Therefore z =
{(2+2^(1/2)^1/2) +i (2-2^(1/2))^(1/2))^(1/2)}^6.

We know 
sqrt[(2+2^(1/2) + (2-2(1/2)] = sqrt4 =2.

Therefore z =
2^6*{[(2+2^(1/2))^(1/2)]/2 +i [(2+2^(1/2))^(1/2)]/2}^6.

z =
2^6 { cos (pi/8) +isin(pi)]^6.

z = 2^6 { cos (6pi/8) +i sin
(6pi/8) } , by De'Moivre's therem.

z = 2^6 {cos 3pi/4 +i
sin 3pi/4}

Threfore modulus of z = |z| = |
2^6{cos(3pi/4)+isin(3pi/4)|

|z| = 2^6 *{(cos(3pi/4))^2
+(sin (3pi/4))^2}

|z| =
2^6.

Therefore modulus of z is 2^6.