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2 Bits of help neede for study Logarithm+Complex number

  1. Sep 28, 2007 #1
    Hey there I'm studying for end of year exams and can't figure out how to do these 2 questions:

    1) Express [tex]\log_{10}(2\sqrt{10}) - \frac{1}{3}\log_{10}0.8 - \log_{10}(\frac{10}{3})[/tex] in the form [tex]c + \log_{10}d[/tex]

    2) Simply finding the cube root of a complex number in polar/cartesian form ie: [tex] (x+yi)^3 = 8i [/tex]
    Last edited: Sep 28, 2007
  2. jcsd
  3. Sep 28, 2007 #2
    for 1) do you remember the properties of a logarithm?

    can't help you with 2) though :/
  4. Sep 28, 2007 #3
    Yeah but How do I raise the whole thing with there being a [tex]\frac{-1}{3}[/tex] Multiplier there?

    Oh I see now, [tex]\sqrt[3]{0.8} = 0.2[/tex]

    AH! I mean, [tex]\sqrt[3]{0.8} = 0.92...[/tex]
    Last edited: Sep 28, 2007
  5. Sep 28, 2007 #4
    yes and then since you have a log - a log you can simplify that, but I don't know how you would get c+log d unless one of those terms simplifies nicely into a base 10 log.
  6. Sep 28, 2007 #5
    the answer scheme says the answer is [tex]\log_{10}3 - \frac{1}{6}[/tex]
  7. Sep 29, 2007 #6
    well expanding everything you'd get:

    but I don't see how they got 1/6..
  8. Sep 29, 2007 #7
    Do it the easy way: translate that Cartesian form that you have into polar form (radius and angle in the form [tex]r e^{i t}[/tex]), find a cube root (straightforward in polar form) and translate that back into Cartesian form.
    Last edited: Sep 29, 2007
  9. Sep 29, 2007 #8

    Gib Z

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    I don't understand the second question, is it asking for us to find the solution of that equation, or to demonstrate that it is true for all complex numbers, which it isn't...

    Anyway, if its the first, its quite straight forward if we take into account 8= 2^3 and i^3 = -i.
  10. Sep 29, 2007 #9
    Then is the [tex]\sqrt[3]{8i} = -2i[/tex]?

    Also i figured out the logarithems one:

    [tex]\log_{10}(2\sqrt{10}) - \frac{1}{3}\log_{10}0.8 - \log_{10}\frac{10}{3}[/tex]
    [tex]= \log_{10}2 + \log_{10}\sqrt{10} - \log_{10}\sqrt[3]{0.8} - \log_{10}10 + \log_{10}3[/tex]
    [tex]= \log_{10}\sqrt[3]{8} - \log_{10}\sqrt[3]{0.8} + \log_{10}\sqrt{10} - \log_{10}10 + \log_{10}3[/tex]
    [tex]= \log_{10}\sqrt[3]{10} + \log_{10}\sqrt{10} - \log_{10}10 + \log_{10}3[/tex]
    [tex]= \frac{1}{3} + \frac{1}{2} - 1 + \log_{10}3[/tex]
    [tex]= \log_{10}3 - \frac{1}{6}[/tex]
  11. Sep 29, 2007 #10

    Gib Z

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    I'm sure if you just cubed both sides you could check that by yourself. But yes it is.
  12. Sep 29, 2007 #11
    It's likely they'll ask a question like, find the [tex]\sqrt[3]{3 + 8i}[/tex] though =/
  13. Sep 29, 2007 #12

    Gib Z

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    These things are very easy if you know the polar form of complex numbers.

    For any complex number a+bi, in polar form it can be written as [tex]\sqrt{a^2+b^2} \exp (i\cdot \arctan (b/a))[/tex]. Euler's formula gives [tex]e^{ix} = \cos x + i \sin x[/tex].

    Using the second formula, try proving the polar form, its just some standard trig.

    Using that, the complex number inside is just 3+8i, which can be written in that polar form. Then, just cube root the [itex]\sqrt{a^2+b^2} [/itex]part on the outside, and for the exponential part just divide the exponent by 3!! Then If you want it back to non polar form, use Euler's formula.
  14. Sep 30, 2007 #13
    Thx Gib_Z I know the process of that way. Is it not doable in cartesian form?
  15. Sep 30, 2007 #14

    Gib Z

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    Unfortunately no, there is 1 method but it is only for square roots (it can be extended to cube and 4th roots only if you know how to solve general polynomials of degree 3 and 4, which is general is a very laborious and difficult process).

    We have the knowledge that the complex field is closed ie Any algebraic operation applied on a complex number will yield another complex number.

    With this knowledge we can conclude that for any complex number, x+iy , its square root will be some other complex number, let it be of the form a+bi. We begin the method thus:

    [tex] \sqrt{ x+ iy} = a+bi[/tex]
    [tex]x+iy = (a+bi)^2 = (a^2-b^2) + 2ab i[/tex]

    By comparing co efficients of real and imaginary parts, we have a pair of simultaneous equations:

    [tex]a^2-b^2 = x[/tex]

    In the second equation, divide through by 2a to find an expression for b.

    Substituting that into the first simultaneous equation gives:
    [tex]a^2 - \frac{y^2}{4a^2} =x[/tex]

    Now multiply through by [itex]4a^2[/itex] and equate the expression to 0.

    Then we have a quadratic equation in a^2. Using the quadratic formula or otherwise, solve for a^2. When we take the square root of that, we have to keep both roots, the positive and the negative, because we are not sure which one is the correct one, or in fact both may me (remember there are always 2 square roots).

    So you have 2 values of a. Each one yields a value of b when put back into either of the simultaneous equations, so there are 2 values of b as well.

    So now you have 4 complex numbers, 2 of them being roots and 2 of them being introduced errors. You only need to check a single one, since the 4 roots come in 2 pairs, that are positive and negatives of each other. It should appear as below:

    [tex] [a+bi, -a-bi] [-a+bi, a-bi][/tex].

    To test one just take the number and square it. If it is correct, so is its negative partner. If its not correct, the other group is correct.

    As you can see, you would probably want to do Polar form instead =D
  16. Sep 30, 2007 #15
    Ah thx gibz
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