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Math Help - Complex Numbers polar

  1. #1
    Member classicstrings's Avatar
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    Complex Numbers polar

    Hello again! A few complex no. questions have got me stuck.

    If z = 1 + i*root3

    Find z bar to the power of -5.

    If w = root2cis(pie/4)

    Find w/z in polar and cartesian form and thus deduce an exact value for cos(pie/12).
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  2. #2
    Junior Member AlvinCY's Avatar
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    Quote Originally Posted by classicstrings View Post
    Hello again! A few complex no. questions have got me stuck.

    If z = 1 + i*root 3

    Find z bar to the power of -5.

    If w = root 2 cis(pi/4)

    Find w/z in polar and cartesian form and thus deduce an exact value for cos(pie/12).

    z = 1 + root 3 i
    z bar = 1 - root3 i

    Converting this to cis form, z bar = 2 cis (-pi/3)

    Using De Moivre's Theorem, (z bar)^5 = (2^5) cis 5*(-pi/3) = 32 cis (pi/3)

    If w = root 2 cis (pi/4)

    w/z in polar form would be (32/ root 2) cis (pi/3 - pi/4) = 16 root 2 cis (pi/12)

    w/z in cartesian form would be:

    w in cartesian form is root2 (cos pi/4 + i sin pi/4) = root 2 [(1/root 2) + i (1/root 2)] = 1 + i

    w/z = (1+i)/(1 + root 3 i)

    Realising the denominator would give you w/z = (1 + root 3)/4 + (1 - root 3)/4 * i

    and thus deduce an exact value for cos(pi/12).

    We know that 16 root 2 cis (pi/12) = 16 root 2 [cos (pi/12) + i sin (pi/12)] = (1 + root 3)/4 + (1 - root 3)/4 * i

    So 16 root 2 * cos (pi/12) = (1 + root 3)/4
    cos (pi/12) = (1 + root 3)/4 divided by 16 root 2...

    Simplify and realise the answer.
    Done!
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  3. #3
    Member classicstrings's Avatar
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    I worked through the problem and still had trouble with the last part.

    I was wondering if someone can check my solutions and perhaps give me a faster more logical way of working through this question. Thanks.

    If z = 1 + i*root3

    i) Find the modulus and argument of z
    ii) Express z^5 in Cartesian form a + ib where a and b are real
    iii) Find z*zbar
    iv) hence, find zbar to the power of negative 5

    If w = root2 cis (π/4)

    i) Find w/z in polar form.

    ii) Express z and w in cartesian form and hence find w/z in cartesian form.

    iii) Use answers from i) and ii) to deduce exact value for cos(π/12)


    The attempt at a solution

    i) Modulus is 2, Argument is π/3

    ii) Using De Moivre's Theorem, 2^5cis(5*(π/3))
    = 32cis(-pie/3)
    Then changing to Cartesian form - 32 (cos(-π/3) + isin(-π/3))
    = 32(0.5 + (root3/2)i)
    = 16 + 16i*root3

    iii) z = 1 + i*root3, zbar = 1 - i*root3
    z*zbar = 4

    iv) Now this part I don't understand the "hence", as in how am I meant to use previous results to get this answer?

    I try - zbar = 2cis(-π/3)

    Then zbar^-5 = Using DM Theorem, (1/32)cis(5π/3)
    = (1/32)cis(-π/3)
    Then convert to cartesian - (1/32)(cos(-π/3) + isin(-π/3))
    = (1/32)(0.5 + iroot3/2)
    = 1/64 + iroot3/64


    If w = root2 cis (π/4)

    i) Find w/z in polar form.
    I found it in cartesian - (1+i)/(1 + root3 i) then realising, gives (1 + root3)/4 + (1-root3)/4 * i

    But how do I change to Polar form? I am also stuck on how to get cos(pie/12) as exact value, even after reading Alvin's comments.
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  4. #4
    Member classicstrings's Avatar
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    Ok I managed to get another part.

    In polar form, it is

    (root2/2)cis(-pie/12)

    Ok after re-reading Alvin's solution I got it myself, thanks!
    Last edited by classicstrings; February 20th 2007 at 02:30 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Ok I managed to get another part.

    In polar form, it is

    (root2/2)cis(-pie/12)

    Ok after re-reading Alvin's solution I got it myself, thanks!
    ARRGH! You know the notation "cis" but you don't know that the Greek letter is spelled "pi?"

    -Dan
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  6. #6
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    Hello, classicstrings!

    Both of you forgot the "minus" . . .


    If .z .= .1 + i√3

    Find z-bar to the power of -5.

    We have: .z .= .2·cis(π/3) .2[cos(π/3) + i·sin(π/3)]

    Then: .z-bar .= .2[cos(-π/3) + i·sin(-π/3)] .= .2[cos(π/3) - i·sin(π/3)]

    Then: .(z-bar)^{-5} .= .(2^{-5})[cos(-5π/3) - i·sin(-5π/3)]

    . . . . . . . . . . . . . . . = .(1/32)[(1/2) - i(√3/2)]

    . . . . . . . . . . . . . . . = .(1/64)(1 - i√3)

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