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Math Help - complex numbers..

  1. #1
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    Post complex numbers..

    i need help with the following questions..

    1. Find all the values of the following:

    a) (1-{\sqrt{3}}i)^{1/3}

    b) (i+1)^{1/2}

    2. Express the complex number z=(1-i)^{10} in the form a+bi

    3. Find the complex number z=a+bi such that z^{3}=2+2i where "a" is less than and equal to 0 and "b" is greater than and equal to 0

    thanks a lot for any help...
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  2. #2
    Member Mentia's Avatar
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    You must be in the same class as "bordas".

    Here is what I said in his post:

    I assume you are using the equation:

     \sqrt[ n]{ z} =  \sqrt[ n]{ r} \left (cos \left (  \frac{ \theta + 2k \pi  }{ n}   \right )  +  isin \left (  \frac{  \theta + 2k \pi }{n }   \right )   \right )

    Theta is the angle Arctan (y/x) where y is the imaginary part and x is the real part of the inner expression (x + iy)^(1/n)

    n is the root, like if you want 3rd root n is 3, and r is the magnitude of the inner expression = (x^2 + y^2) ^ (1/2)

    plug in each of k = 0, 1, 2

    and see what happens.



    Also, dont forget De Moivre's formula to solve #2:

    z^n = r^n * ( cos (n*T) + i sin (n*T) )

    where T is theta, given by the arctan, and r is the magnitude, the same things as above.
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  3. #3
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    Quote Originally Posted by Mentia View Post
    You must be in the same class as "bordas".

    Here is what I said in his post:

    I assume you are using the equation:

     \sqrt[ n]{ z} = \sqrt[ n]{ r} \left (cos \left ( \frac{ \theta + 2k \pi }{ n} \right ) + isin \left ( \frac{ \theta + 2k \pi }{n } \right ) \right )

    Theta is the angle Arctan (y/x) where y is the imaginary part and x is the real part of the inner expression (x + iy)^(1/n)

    n is the root, like if you want 3rd root n is 3, and r is the magnitude of the inner expression = (x^2 + y^2) ^ (1/2)

    plug in each of k = 0, 1, 2

    and see what happens.



    Also, dont forget De Moivre's formula to solve #2:

    z^n = r^n * ( cos (n*T) + i sin (n*T) )

    where T is theta, given by the arctan, and r is the magnitude, the same things as above.

    ok for #2 i got -32i and for #3 i got 2cos(pi/12)+2isin(pi/12) are those answers right??? and i am still stuck on #1..any help
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  4. #4
    Member Mentia's Avatar
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    Your #2 looks good!

    I don't think your #3 is correct though. Lets take a look:

    ArcTan(2/2) = Pi/4 = Theta

    r = (2^2+2^2)^(1/2) = 8^(1/2)

    n = 3

    then z = 8^(1/6) * ( cos[ (Pi/4 + 2*Pi*k) / n ] + i*sin[ (Pi/4 + 2*Pi*k) / n ])

    plug in 0, 1 and 2 for k and you get the answers (remember, you plug in values for k such that k = 0, 1, ..., n-1:

    I am going to give you the decimal solutions, you can figure out the exact if you wish.

    k = 0 -> z = 1.366025 + i*.366025
    k = 1 -> z = -1 + i
    k = 2 -> -.366025 - i*1.366025

    so the answer to your question #3 is the root where k = 1:

    z = -1 + i

    raise that to the 3rd power and you'll find z^3 = 2 + 2i as expected.


    As far as #1 is concerned, you do the same thing as I just did for #3, you find all the roots by plugging in all the k values. Note that k = 0, 1, ..., n - 1.

    So if you want the 2nd root you use k = 0 or 1
    if you want the 3rd root you use k = 0, 1, or 2
    4th root k = 0, 1, 2, or 3
    etc.
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  5. #5
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    Quote Originally Posted by Mentia View Post
    Your #2 looks good!

    I don't think your #3 is correct though. Lets take a look:

    ArcTan(2/2) = Pi/4 = Theta

    r = (2^2+2^2)^(1/2) = 8^(1/2)

    n = 3

    then z = 8^(1/6) * ( cos[ (Pi/4 + 2*Pi*k) / n ] + i*sin[ (Pi/4 + 2*Pi*k) / n ])

    plug in 0, 1 and 2 for k and you get the answers (remember, you plug in values for k such that k = 0, 1, ..., n-1:

    I am going to give you the decimal solutions, you can figure out the exact if you wish.

    k = 0 -> z = 1.366025 + i*.366025
    k = 1 -> z = -1 + i
    k = 2 -> -.366025 - i*1.366025

    so the answer to your question #3 is the root where k = 1:

    z = -1 + i

    raise that to the 3rd power and you'll find z^3 = 2 + 2i as expected.


    As far as #1 is concerned, you do the same thing as I just did for #3, you find all the roots by plugging in all the k values. Note that k = 0, 1, ..., n - 1.

    So if you want the 2nd root you use k = 0 or 1
    if you want the 3rd root you use k = 0, 1, or 2
    4th root k = 0, 1, 2, or 3
    etc.
    how did u figure out that k = 1??
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  6. #6
    Member Mentia's Avatar
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    k = 1 gives the only z where the real part is negative and the imaginary part is positive like it asks.

    "Find the complex number z=a+bi such that where "a" is less than and equal to 0 and "b" is greater than and equal to 0"

    a is the real part, b is the imaginary part

    Notice you can write -1 + i as -1 + 1*i -> a = -1, b = +1

    That help?
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  7. #7
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    Quote Originally Posted by Mentia View Post
    k = 1 gives the only z where the real part is negative and the imaginary part is positive like it asks.

    "Find the complex number z=a+bi such that where "a" is less than and equal to 0 and "b" is greater than and equal to 0"

    a is the real part, b is the imaginary part

    Notice you can write -1 + i as -1 + 1*i -> a = -1, b = +1

    That help?
    yeah that helps alot..thanks
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