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Math Help - modulus-argument form

  1. #1
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    modulus-argument form

    Let z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}

    (a) express z in modulus-argument form

    (b) express z in real-imaginary form

    First Question: Do I need to expand out the top and bottom parts before doing anything else?

    Edit: This is what I got for (b):
    expanding the denominator out, we get (-2 + 2i)
    multiplying by \frac{(-2 - 2i)}{(-2 - 2i)} gives \frac{(4i - 4)(1 + \sqrt{3i})^3}{8} = \frac{(i-1)(1 + \sqrt{3i})^3}{2}
    expanding out (1 + \sqrt{3i})^3 gives:
    [1 + 2\sqrt{3i} + 3i](1 + \sqrt{3i}) = 1 + 3\sqrt{3i} + 9i + 3i\sqrt{3i}
    3i\sqrt{3i} = 3\sqrt{3i^3} = -3\sqrt{3i} thus:
    = 1 + 3\sqrt{3i} + 9i - 3\sqrt{3i} =  1 + 9i
    hence:
    z = \frac{(i -1)(1 + 9i)}{2}
    z =  \frac{(-10 -8i)}{2}
    z = -5 - 4i

    correct? yes/no?

    Any help for part (a) greatly appreciated!



    FURTHER EDIT:
    I've been through Moo's workings, adding in the cube power he missed and got the following:
    (using Moo's notation of x, y, & t)
    x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{2}=2 e^{-i \tfrac \pi {2}}}
    {y=2 e^{i \tfrac \pi 3}}\implies \boxed{y^{3}=8 e^{i \pi} = -8}
    t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{3}=2 \sqrt{2} e^{i \tfrac{{3}\pi} 4}}

    which gives:
    <br />
z=\frac{2 e^{-i \tfrac \pi 2} \cdot -8}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}} \implies<br />
-4\sqrt{2} \cdot \frac{e^{-i \tfrac \pi 2}}{e^{i \tfrac{3\pi} 4}} \implies -4\sqrt{2} \cdot e^{i (\tfrac {-\pi} {2} - \tfrac {3\pi} {4})} \implies 4\sqrt{2} \cdot e^{-i \tfrac {5\pi} {4}}<br />

    <br />
z= -4\sqrt{2} (cos \tfrac {-5\pi} {4} + isin \tfrac {-5\pi} {4}) \implies-4\sqrt{2}(-\tfrac {1} {\sqrt{2}} + i \tfrac {1} {\sqrt{2}})
    <br />
z= 4 - 4i

    which is ever-so-slightly different to what I got by expanding it all out.

    What the hell am I doing wrong here?! I've spent 1/2 the morning going over this and just can't for the life of me see where I've gone wrong. Any help desperately appreciated!
    Last edited by Dr Zoidburg; October 2nd 2008 at 07:41 PM.
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  2. #2
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    I'm not sure what modulus-argument form is, but in order to express this number as a + bi, I would multiply by \frac{(1 - i)^3}{(1 - i)^3} in order to get a real denominator. Then you can multiply out the top part and simplify it and everything will work out. It's a bit tedious, but I can't think of an easier way of doing it.
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  3. #3
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    That's what I figured for the 2nd part as well. Though it's a little easier to expand the bottom out first, as that gives (-2 + 2i) and then muliply both by (-2 - 2i) which gives us 8 as the demoninator.


    I hope.
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  4. #4
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    Hello,

    Yes, it gives 8. But you don't need to do it this way, you can keep all the original stuff.

    Remember that e^{ix}=\cos(x)+\sin(x) i

    z=\frac{(1-i)^2 (1+\sqrt{3} i)}{(1+i)^3}=\frac{x^2 \cdot y}{t^3}
    In order to have the modulus argument form, you have to divide each complex part by its modulus.

    x=1-i \implies |x|=\sqrt{1+1}=\sqrt{2}
    \frac{x}{\sqrt{2}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i=e^{-i \tfrac \pi 4}

    x=\sqrt{2} e^{-i \tfrac \pi 4} \implies \boxed{x^{\color{red}2}=2 e^{-i \tfrac \pi {\color{red}2}}}



    y=1+\sqrt{3} i \implies |y|=\sqrt{1+(\sqrt{3})^2}=2
    So we have \frac{y}{2}=\frac{1+\sqrt{3} i}{2}=\frac 12+\frac{\sqrt{3}}{2} i=e^{i \tfrac \pi 3}

    \boxed{y=2 e^{i \tfrac \pi 3}}



    t=1+i \implies |t|=\sqrt{1+1}=\sqrt{2}
    \frac{t}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sq  rt{2}} i=e^{i \tfrac \pi 4}

    t=\sqrt{2} e^{i \tfrac \pi 4} \implies \boxed{t^{\color{red}3}=2 \sqrt{2} e^{i \tfrac{{\color{red}3}\pi} 4}}


    z=\frac{2 e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{2 \sqrt{2} e^{i \tfrac{3\pi} 4}}

    z=\frac{1}{\sqrt{2}} \cdot \frac{e^{-i \tfrac \pi 2} \cdot e^{i \tfrac \pi 3}}{e^{i \tfrac{3\pi} 4}}

    \frac{1}{\sqrt{2}} is the modulus. Now, use basic exponent rules to find the argument
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  5. #5
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    Moo, I just noticed that you wrote the equation as
    z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}
    you missed out the ^3 power up top:
    z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}

    does this change the answer?
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  6. #6
    Moo
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    Quote Originally Posted by Dr Zoidburg View Post
    Moo, I just noticed that you wrote the equation as
    z=\frac{(1-i)^2(1+\sqrt{3i})}{(1+i)^3}
    you missed out the ^3 power up top:
    z=\frac{(1-i)^2(1+\sqrt{3i})^3}{(1+i)^3}

    does this change the answer?
    Of course it changes it !
    But the reasoning is completely the same. Please try to do it
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  7. #7
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    I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...
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  8. #8
    Moo
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    Quote Originally Posted by Dr Zoidburg View Post
    I have! It's in the OP! I just hope it's correct, cause I sent the assignment off earlier today...
    Oh sorry, I didn't see that...

    Well, I don't see any mistake, after checking twice. However, given my state, don't take my word for granted at 100%.
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  9. #9
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    I'll just hope it's either correct or the marker's in the same state you're in when s/he marks it!
    thanks for the help
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