Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

write the Complex number in Polar Coordinates?

1) Evaluate (2+3i)^1/2

2) e^(2+pi/2i)

I know you use De moivres Theorem but when I get tan inverse of 3/2 it gives me a decimal point. So I obviously have to do something different but what?

Any help would be great thanks.

2 Answers

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  • 7 years ago

    1)

    √(2 + 3i)

    go polar

    x + iy = ρ(cos θ + i sin θ)

    where

    ρ = √(x^2 + y^2)

    θ = arctan(y/x)

    √(2 + 3i) = x + iy

    square both sides

    2 + 3i = x^2 + i^2 y^2 + 2ixy

    2 + 3i = (x^2 - y^2) + (2xy) i

    must be

    x^2 - y^2 = 2

    2xy = 3

    from the second equation

    y = 3/(2x)

    substitute in the first

    x^2 - 9/(4x^2) = 2

    set x^2 = w

    4w - 9/w = 8

    4w^2 - 8w - 9 = 0

    w = 1 - √13/2 ∨ w = √13/2 + 1

    drop the negative root

    x = √(√13/2 + 1)

    y = 3/(2√(√13/2 + 1)) = √(√13/2 - 1)

    √(2 + 3i) = √(√13/2 + 1) + i √(√13/2 - 1)

    2)

    e^(2+π i/2) = e^2 e^π i/2 = e^2 i

  • (2 + 3i)^(1/2)

    a = 2

    b = 3

    sqrt(a^2 + b^2) = sqrt(2^2 + 3^2) = sqrt(13)

    (sqrt(13) * (2/sqrt(13) + i * 3/sqrt(13)))^(1/2) =>

    13^(1/4) * (cos(t) + i * sin(t))^(1/2)

    cos(t) = 2/sqrt(13)

    t = 0.98279372324732906798571061101467 + 2pi * k radians

    13^(1/4) * (cos(0.983 + 2pi * k) + i * sin(0.983 + 2pi * k))^(1/2)

    13^(1/4) * (cos(0.4914 + pi * k) + i * sin(0.4914 + pi * k))

    13^(1/4) * (cos(0.4914) + i * sin(0.4914)) and 13^(1/4) * (cos(3.633) + i * sin(3.633))

    (13^(1/4) , 0.4914)

    (13^(1/4) , 3.633)

    e^(2 + (pi/2) * i) =>

    e^(2) * (cos(pi/2) + i * sin(pi/2)) =>

    e^(2) * (0 + i * 1) =>

    e^(2) * i

    (e^(2) , pi/2)

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