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    I'm confused with the following:

    Write -5(cos[pi/4] - i.sin[pi/4]) in the form a + ib, a, b are members of the reals.

    I understand the methods to use, but it is another aspect that is confusing me.

    -5 represents r which is equivalent to the modulus of the complex no (say z). However, surely the modulus of z will always be postive, since it is equal to the positive square root of (a^2 + b^2) and since a and b are reals.

    Or, is that when dealing with the modulus of complex no's we can also take the negative square root to be the modulus?!

    If someone could clear up my confusiong that would be great. Thanks
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    Ok

    -5(cos theta - i sin theta) = 5(i sin theta - cos theta)

    So basically r = 5,
    and the principle argument is Below the real axis? (b < 0) I'm not sure
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    The modulus is always >= 0.

    Modulus of product = Product of the moduluses: |zw| = |z| * |w|.

    In your case,

    |-5 * (cos[pi/4] - i.sin[pi/4])|
    = |-5| * |cos[pi/4] - i.sin[pi/4]|
    = |-5| * |cos[-pi/4] + i.sin[-pi/4]|
    = 5 * 1
    = 5.
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    Thank you both very much. Could one of you help me with:

    Given that z=5 - 12i express sqrt(z) in the form a + ib where a,b, is a member of the reals

    I was not sure how to deal with the sqrt of z (and therefore sqrt of [5 - 12i]). Can I use bionomial theorem even though it contains imaginary no's?
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    (Original post by Jonny W)
    The modulus is always >= 0.

    Modulus of product = Product of the moduluses: |zw| = |z| * |w|.

    In your case,

    |-5 * (cos[pi/4] - i.sin[pi/4])|
    = |-5| * |cos[pi/4] - i.sin[pi/4]|
    = |-5| * |cos[-pi/4] + i.sin[-pi/4]|
    = 5 * 1
    = 5.
    exactly what I was thinking.

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    (Original post by Hoofbeat)
    Thank you both very much. Could one of you help me with:

    Given that z=5 - 12i express sqrt(z) in the form a + ib where a,b, is a member of the reals

    I was not sure how to deal with the sqrt of z (and therefore sqrt of [5 - 12i]). Can I use bionomial theorem even though it contains imaginary no's?

    Let a+ib = (5-12i)^(1/2). It follows a^2-b^2 = 5 and ab = -6 by squaring. This gives a^2 - 36/a^2 = 5. This gives a^4 - 5a^2 - 36 = 0. i.e (a^2-9)(a^2+4) = 0. So a = +/- 3, b = -/+ 2.
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    (Original post by Hoofbeat)
    Given that z=5 - 12i express sqrt(z) in the form a + ib where a,b, is a member of the reals
    Let a + bi equal sqrt(z)

    Then (a + bi)^2 = z
    a^2 + 2abi + (bi)^2 = z
    a^2 - b^2 + 2abi = 5 - 12i
    Hence a^2 - b^2 = 5 and 2ab = -12

    Use a = -12/2b or b = -12/2a to find the values of a and b. You'll turn up two answers (which will be conjugates).
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    Thanks Squishy & Beauford - It's so easy (or I'm v.thick!). Grrr wish my mind hadn't gone to sleep at the moment!
 
 
 
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Updated: August 17, 2004
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