Final answer:
To find the complex fifth roots of −16√2+16√2i in exponential form, we can use De Moivre's theorem. The roots are: Root 1: 2^(1/5)(cos(-π/20) + i sin(-π/20)), Root 2: 2^(1/5)(cos(7π/20) + i sin(7π/20)), Root 3: 2^(1/5)(cos(15π/20) + i sin(15π/20)), Root 4: 2^(1/5)(cos(23π/20) + i sin(23π/20)), Root 5: 2^(1/5)(cos(31π/20) + i sin(31π/20)).
Step-by-step explanation:
To find the complex fifth roots of −16√2+16√2i in exponential form, we can use De Moivre's theorem. De Moivre's theorem states that for any complex number z = r(cos θ + i sin θ), its nth power can be expressed as z^n = r^n(cos nθ + i sin nθ).
- First, we need to find the magnitude (r) and argument (θ) of the complex number. The magnitude is given by |z| = √(Re^2 + Im^2), where Re represents the real part and Im represents the imaginary part. In this case, |z| = √((-16√2)^2 + (16√2)^2) = √(512 + 512) = √1024 = 32.
- The argument can be determined using the formula θ = atan(Im/Re). In this case, θ = atan((16√2)/(-16√2)) = atan(-1) = -π/4.
- Now, we can express the complex number in exponential form: -16√2+16√2i = 32(cos(-π/4) + i sin(-π/4)).
- To find the fifth roots, we need to find the complex numbers whose fifth power is equal to the given complex number. To do this, we can express the complex number in polar form as 32(cos(-π/4) + i sin(-π/4)).
- Now, we can apply De Moivre's theorem by taking the fifth root of the magnitude and dividing the argument by 5: (32^(1/5))(cos(-π/4)/5 + i sin(-π/4)/5).
By simplifying the expression, we get the complex fifth roots of −16√2+16√2i in exponential form:
Root 1: 2^(1/5)(cos(-π/20) + i sin(-π/20))
Root 2: 2^(1/5)(cos(7π/20) + i sin(7π/20))
Root 3: 2^(1/5)(cos(15π/20) + i sin(15π/20))
Root 4: 2^(1/5)(cos(23π/20) + i sin(23π/20))
Root 5: 2^(1/5)(cos(31π/20) + i sin(31π/20))