How can I find the argument (angle) of -81i?

This is a portion from the problem: Find Solution strictly between 90 degrees and 180 degrees for z^(2) = -81i. I know the modulus is 81, but I don't know how to find the argument when only 81i is given.

Thanks.

7 Answers

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  • 1 month ago

    z^2=-81i=>

    z^2=81(-i)=>

    z^2=81e^[(2kpi-pi/2)i]=>

    z=9e^[(kpi-pi/4)i]

    z0=9e^[(-pi/4)i], k=0

    z1=9e^[(3pi/4)i], k=1

    =>

    the argument in (pi/2, pi) is 3pi/4.

  • 1 month ago

    -81i = 81[cos(270° + 360n°) + isin(270° + 360n°)]

    so, √-81i = 9[cos(270° + 360n°) + isin(270° + 360n°)] ¹/² 

    i.e. √-81i = 9[cos(135° + 180n°) + isin(135° + 180n°)]

    With n = 0 we have:

    9[cos135° + isin135°] => 9[-√2/2 + i√2/2]

    or, (9√2/2)[-1 + i]

    With n = 1 we have:

    9[cos315° + isin315°] => 9[√2/2 - i√2/2]

    or, (9√2/2)[1 - i]

    Hence,  ±(9√2/2)[1 - i]

    Note, squaring will give -81i

    :)>

     

  • Dixon
    Lv 7
    1 month ago

    Doesn't anyone use polar coordinates???

    (A ∠a)  x (B ∠b) = (A x B) ∠(a + b)

    So the primary solution of √(A ∠a) is just √A ∠a/2

    And in general, when taking the nth root of a complex number there are n solutions. These solutions are all the same length (modulus) and are spaced out regularly at angles of 360°/n like spokes of a wheel, centered around the primary solution.

    So there will be two solutions which will be in opposite directions, and in your head you can just look at z² = -81i and go

    z² = -81i

    z² = 81 ∠270°

    z = 9 ∠(270°/2) and another solution at ± 180° from that

    ie

    z = 9 ∠135 , 9 ∠-45°

  • ted s
    Lv 7
    1 month ago

    - 81 i = 81  ( - i )  = 81 ( cos 3π / 2 + i sin 3π / 2 ) ===> no angle in [ 90° , 180° ]...but you want z² = - 81 i = 81 ( cos Θ+ i sin Θ , where Θ = 270 + n 360 which implies z = 9 ( cos Θ/2 + i sin Θ/2 ) , Θ / 2 = 135 + n 180 ....135° is in [ 90° , 180° ]

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  • 1 month ago

    Pure imaginary values are on the "y-axis", so positive multiples of i all have angle pi/2 and  negative multiples of i all have have angle 3*pi/2 or -pi/2.

    You kinda need to just know that since the arctan(y/x) formula blows up there.

    I tend to use arccos(x/r) instead in computer programs for that reason, since I usually need to know both r and θ, (The sign of θ is negative, -arccos(x/r), when y < 0.)

  • z^2 = -81i

    z^2 = 81 * i^(3 + 4k)

    z = 9 * i^(1.5 + 2k)

    z = 9 * i^(1.5) , 9 * i^(3.5)

    i^(1.5) =>

    (0 + i)^(1.5) =>

    (cos(pi/2) + i * sin(pi/2))^(1.5) =>

    cos(1.5 * pi/2) + i * sin(1.5 * pi/2) =>

    cos(3pi/4) + i * sin(3pi/4)

    i^(3.5) =>

    (0 + i)^(3.5) =>

    (cos(pi/2) + i * sin(pi/2))^(7/2) =>

    cos(7pi/4) + i * sin(7pi/4)

    3pi/4 => 135 degrees

    7pi/4 => 315 degrees

    90 < t < 180

    t = 135 degrees

    The modulus is 9

  • 1 month ago

    Let z = a + bi, where a and b are real and i^2 = -1, so we have:

    (a + bi)^2 = -81i

    a^2 + 2abi + b^2i^2 + 81i = 0

    a^2 - b^2 + 2abi + 81i = 0

    Therefore:

    a^2 - b^2 = 0, so a^2 = b^2, so a = +/- b

    2ab + 81 = 0, so 2ab = -81, so ab = -81/2 = -40.5

    So +/- b * b = -40.5

    b^2 = +/- 40.5

    Since b is real, b^2 can't be negative, so we have:

    b^2 = 40.5

    b = +/- sqrt(40.5)

    If b = sqrt(40.5) then a = -sqrt(40.5), and vice versa.

    So z = sqrt(40.5) - sqrt(-40.5) or z = -sqrt(40.5) + sqrt(-40.5)

    Note that sqrt(40.5) =~ 6.3639610306789277196075992589436

    And sqrt(-40.5) =~ 6.3639610306789277196075992589436 i

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