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    How would I find the modulus and principle argument of this monster?

    \frac{4+4i}{(\sqrt{3}+i)^5}

    :hmmmm:
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    (Original post by RamocitoMorales)
    How would I find the modulus and principle argument of this monster?

    \frac{4+4i}{(\sqrt{3}+i)^5}

    :hmmmm:
    Work out the modulus and principle argument for the component parts and then add/subtract multiply/divide as necessary.
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    (Original post by ghostwalker)
    Work out the modulus and principle argument for the component parts and then add/subtract multiply/divide as necessary.
    For the top part, I find that the modulus is 4\sqrt{2} and the argument is \frac{\pi}{4}. For the bottom part, I find that the modulus is 32 and the argument is \frac{5\pi}{6}.

    So what do I do? :puppyeyes:

    I tried combining the two fractions together, but I ended up with something stupid,

    z=4+4i=4\sqrt{2}(cos\frac{\pi}{4  }+isin\frac{\pi}{4})

    w=(\sqrt{3}+i)^5=32(cos\frac{5\p  i}{6}+isin\frac{5\pi}{6})

    \frac{z}{w}=\frac{128-64i+64\sqrt{3}i}{1024}

    :cry2:
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    (Original post by RamocitoMorales)
    For the top part, I find that the modulus is 4\sqrt{2} and the argument is \frac{\pi}{4}. For the bottom part, I find that the modulus is 32 and the argument is \frac{5\pi}{6}.

    So what do I do? :puppyeyes:

    I tried combining the two fractions together, but I ended up with something stupid,

    z=4+4i=4\sqrt{2}(cos\frac{\pi}{4  }+isin\frac{\pi}{4})

    w=(\sqrt{3}+i)^5=32(cos\frac{5\p  i}{6}+isin\frac{5\pi}{6})

    \frac{z}{w}=\frac{128-64i+64\sqrt{3}i}{1024}

    :cry2:
    When you divide two complex numbers, one by the other, all you need do is divide their modulii, and subtract their arguments (the right way round of course).
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    (Original post by ghostwalker)
    When you divide two complex numbers, one by the other, all you need do is divide their modulii, and subtract their arguments (the right way round of course).
    It worked!

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    (Original post by RamocitoMorales)
    It worked!
    Your previous answer should work out correctly as well, but I can't be arsed to check it.
 
 
 
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