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

neela | High School Teacher | (Level 3) Valedictorian

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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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We can use the following: we'll write sqrt(2+sqrt2) as 2*cos (pi/4) and sqrt(2-sqrt2) = 2*sin (pi/4).

We know that:

cos pi/8 = sqrt {[1+cos(pi/4)]/2}

cos pi/4 = sqrt2/2

cos pi/8 = sqrt {[1 + sqrt2/2]/2}

cos pi/8 = [sqrt (2+sqrt2)]/2

2*cos pi/8 = sqrt (2+sqrt2)

2*sin pi/8 = sqrt (2-sqrt2)

We'll re-write z:

z = [2*cos (pi/8) + i*2*sin (pi/8)]^6

We'll factorize by 2:

z = {2[cos (pi/8) + i*sin (pi/8)]}^6

z = 2^6*[cos (pi/8) + i*sin (pi/8)]^6 (1)

We also know that when we put a complex number in a trigonometric form, we'll get the expression:

z = |z|*(cos a + i*sin a) (2)

If we make a comparison between the expressions (1) and (2), we notice that expression (1) is the trigonometric form of the complex number, whose modulus is |z| = 2^6

|z| = 2^6