I need the answers for this question please : show that 2^(1/4) (1-i) is a fourth root of -2?
There is a problem here.
If you take 2^(1/4) (1-i) to the 4th power you get:
{2^(1/4) (1-i)}^4 = 2*(1-i)^4
And (1 – i)^4 = {(1 – i)^2}^2 = {1 - 2i + (i)^2}^2 = (1 – 2i – 1)}^2 = {-2i}^2 = -4
i.e. (1 – i)^4 = -4, not -1 as needed.
Your expression would need to be 2^(1/4) (1-i)/√2 to give (-2) when taken to the fourth power.
Assuming the typo ...
Let z = -2
i.e. z = -2(cos(2nπ) + i.sin(2nπ)), n = 0,1,2,3, ...
or, z = 2(cos(π + 2nπ) + i.sin(π + 2nπ)), n = 0,1,2,3, ...
z = 2.e^(i.(π + 2nπ)) ---- Euler’s formula
Taking the fourth root of z, as Z4 = z^(1/4)
Z4 = (2)^(1/4).e^(i(π/4 + nπ/2))
Z4 = (2)^(1/4). (cos(π/4 + nπ/2) + i.sin(π/4 + nπ/2))
Letting n = 0,1,2,3
Z41 = (2)^(1/4). (cos(π/4) + i.sin(π/4)) = (2)^(1/4). (1/√2 + i. 1/√2)
Z41 = (2)^(1/4)(1 + i)/√2
Z42 = (2)^(1/4). (cos(π/4 + π/2) + i.sin(π/4 + π/2))
Z42 = (2)^(1/4). (cos(3π/4) + i.sin(3π/4)) = (2)^(1/4). (-1/√2 + i. 1/√2)
Z42 = (2)^(1/4). (-1 + i)/√2
Z43 = (2)^(1/4). (cos(π/4 + π) + i.sin(π/4 + π))
Z43 = (2)^(1/4). (cos(5π/4) + i.sin(5π/4)) = (2)^(1/4). (-1/√2 – i. 1/√2)
Z43 = (2)^(1/4). (-1 – i)/√2
Z44 = (2)^(1/4). (cos(π/4 + 3π/2) + i.sin(π/4 + 3π/2))
Z44 = (2)^(1/4). (cos(7π/4) + i.sin(7π/4)) = (2)^(1/4). (1/√2 – i. 1/√2)
Z44 = (2)^(1/4). (1 – i)/√2
The final expression, Z44, shows that the 4th fourth root of (-2) is (2)^(1/4). (1 – i)/√2, as required