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Will asked in Science & MathematicsMathematics · 11 months ago

What is (√2 + 1i)^5 using DeMoivres Theorem?

Update:

Question on my sample test. (√2 + 1i)^5 using DeMoivres Theorem

3 Answers

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  • 11 months ago
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    (a + bi)ᶜ

    = (r eⁱᶿ)ᶜ where r = √(a²+b²) and θ = atan2(b,a)

    = rᶜ (eⁱᶿ)ᶜ

    = rᶜ eⁱ⁽ᶜᶿ⁾

    = (a²+b²)^(c/2) cis(cθ)

    = (a²+b²)^(c/2) cos(cθ) + i (a²+b²)^(c/2) sin(cθ)

    (√2 + 1i)^5

    = (√2²+1²)^(5/2) cos(5*atan2(1,√2)) + i (√2²+1²)^(5/2) sin(5*atan2(1,√2))

    = 9√3 cos(5*atan2(1,√2)) + i 9√3 sin(5*atan2(1,√2))

    = -√242 + i

    = -11√2 + i

    Confirm if required by expanding (√2 + i)^5 using binomial theorem:

    i^5 + 5√2 i^4 + 20 i^3 + 20√2 i^2 + 20 i + 4√2

    = i + 5√2 - 20i - 20√2 + 20i + 4√2

    = -11√2 + i ✓

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  • sqrt(2) + i = R * cos(t) + R * sin(t) * i

    R * cos(t) = sqrt(2)

    R * sin(t) = 1

    R^2 * (cos(t)^2 + sin(t)^2) = 2 + 1

    R^2 = 3

    R = sqrt(3)

    (sqrt(3) * (sqrt(2/3) + i/sqrt(3)))^5 =>

    3^(5/2) * (sqrt(2/3) + i/sqrt(3))^5 =>

    9 * sqrt(3) * (cos(t) + i * sin(t))^5

    (cos(t) + i * sin(t))^5 =>

    cos(5t) + i * sin(5t)

    cos(5t) =>

    cos(4t + t) =>

    cos(4t)cos(t) - sin(4t)sin(t) =>

    (cos(2t)^2 - sin(2t)^2) * cos(t) - 2sin(2t)cos(2t)sin(t) =>

    (2 * cos(2t)^2 - 1) * cos(t) - 4sin(t)^2 * cos(t) * (cos(t)^2 - sin(t)^2) =>

    (2 * (cos(t)^2 - sin(t)^2)^2 - 1) * cos(t) - 4 * sin(t)^2 * cos(t) * (cos(t)^2 - sin(t)^2) =>

    (2 * ((2/3) - (1/3))^2 - 1) * sqrt(2/3) - 4 * (1/3) * sqrt(2/3) * ((2/3) - (1/3)) =>

    (2 * (1/3)^2 - 1) * sqrt(2/3) - (4/3) * sqrt(2/3) * (1/3) =>

    (2 * (1/9) - 1) * sqrt(2/3) - (4/9) * sqrt(2/3) =>

    (-7/9) * sqrt(2/3) - (4/9) * sqrt(2/3) =>

    (-11/9) * sqrt(2/3)

    sin(5t) =>

    sin(4t + t) =>

    sin(4t)cos(t) + sin(t)cos(4t) =>

    2sin(2t)cos(2t)cos(t) + sin(t) * (cos(2t)^2 - sin(2t)^2) =>

    4sin(t)cos(t)^2 * (cos(t)^2 - sin(t)^2) + sin(t) * (2 * cos(2t)^2 - 1) =>

    4 * sin(t) * cos(t)^2 * (cos(t)^2 - sin(t)^2) + sin(t) * (2 * (2 * cos(t)^2 - 1)^2 - 1) =>

    4 * sqrt(1/3) * (2/3) * (2/3 - 1/3) + sqrt(1/3) * (2 * (2 * (2/3) - 1)^2 - 1) =>

    sqrt(1/3) * ((8/3) * (1/3) + (2 * (4/3 - 1)^2 - 1)) =>

    sqrt(1/3) * (8/9 + (2 * (1/3)^2 - 1)) =>

    sqrt(1/3) * (8/9 + (2/9 - 1)) =>

    sqrt(1/3) * (8/9 + 2/9 - 9/9) =>

    sqrt(1/3) * (1/9)

    9 * sqrt(3) * (cos(t) + i * sin(t))^5 =>

    9 * sqrt(3) * (cos(5t) + i * sin(5t)) =>

    9 * sqrt(3) * ((-11/9) * sqrt(2/3) + i * sqrt(1/3) * (1/9)) =>

    -11 * 9 * sqrt(3) * sqrt(2) / (9 * sqrt(3)) + i * 9 * sqrt(3) / (sqrt(3) * 9) =>

    -11 * sqrt(2) + i

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  • Vaman
    Lv 7
    11 months ago

    tet sqrt 2= r cos a, 1 = r sin a, r^2= 2+1=3. r= sqrt 3.

    tan a= 1/sqrt 2. this gives a= 35.26 .

    The value sqrt 2 + 1= sqrt 3 exp i a. Take to power 5. (sqrt 3)^5= 9 sqrt 3. 5a=176.3. The answer is 9 sqrt 3 exp (i 176.3)

    = 9 sqrt 3 ( cos 176.3 + i sin 176.3)

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