Find ((1/ square root of 2)+(i/square root of 2)) raised ot the power of 8?

Answer 1

z=(1/√2+i/√2)^8

Mod z = √((1/√2)^2 + (1/√2)^2) = 1
Arg(z) = pi/4

Use deMoivre's Theorem

Mod (z^8) = [Mod(z)]^8 = 1^8 = 1

Arg(z^8) = 8*Arg(z) = 2pi --- equivalent to 0

The answer is 1

z is actually sqrt(i)
z^2 = i
z^4 = i^2 = -1
z^8 = (-1)^2 = 1

Answer 2

First combine the fractions.
1/sqrt2 + i/sqrt2 = (1+i)/sqrt2
sqrt2 = 2^(1/2)
2^((1/2)^8) = 2^4 = 16
The denominator will be 16
to find the numerator you can use Pascal's triangle
1 (a+b)^0
1 1 (a+b)^1
1 2 1 (a+b)^2
1 3 3 1 ...^3
1 4 6 4 1 ...^4
1 5 10 10 5 1 ...^5
1 6 15 20 15 6 1 ...^6
1 7 21 35 35 21 7 1 ...^7
1 8 28 56 70 56 28 8 1 ...^8
This gives you the coefficients of the expanded binomial

1 +8i -28 -56i + 70 + 56i -28 - 8i +1
[i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, etc]

combining like terms gives you
72 - 56 = 16 for the numerator
16/16 = 1

Answer 3

On the complex plane, 1 + i is a point with coordinates (1,1). Its distance from the origin is √2. So (1/√2)(1,1) is on the unit circle, and in polar terms it is (1, 45°). DeMoivre's theorem says 8th power is (1)^8 [ cos(45°•8) + i sin (45°•8) = cos 0 + i sin 0 = 1

Answer 4

((1/sqrt(2) + i/sqrt(2))^8
((1 + i)/sqrt(2))^8
((1 + i)^8)/(2^(8/2))
((1 + i)^8)/(2^4)
((1 + i)^8)/64
(1 + i)(1 + i)(1 + i)(1 + i)(1 + i)(1 + i)(1 + i)(1 + i)
(1 + 2i - 1)(1 + 2i - 1)(1 + 2i - 1)(1 + 2i - 1)
(2i)^4
64((-1)^2)
64/64
1

((1/sqrt(2)) + (i/sqrt(2))^8 = 1

Answer 5

[(1/sqrt2)+(1/sqrt2)}^8
= {(1+1)/sqrt2}^8
=(2/sqrt2)^8
=2^8/sqrt2^8
=2^8/2^4
=2^(8-4)
=2^4
=16 ans