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Math Help - [SOLVED] Rectangular form of a complex number

  1. #1
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    [SOLVED] Rectangular form of a complex number

    You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

    The problem is to put
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}
    in rectangular form.

    I'm kinda stuck on this one.

    Before I mention what I've tried, let me say what the book's answer for this one is:
    (2 - \sqrt{3})^{12}

    My plan of attack was to write \sqrt{3 - i} in rectangular form, giving me
    \sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )
    (By the way, this approach doesn't use that there are actually two values for \sqrt{3 - i}. I'm not sure which one to pick, if it even makes a difference.)

    Now that I've got that, then I can rewrite the original problem:
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =  \left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}

    Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

    And it certainly isn't the nice form (2 - \sqrt{3})^{12}.

    Any thoughts as to how to approach this? Thanks!

    -Dan
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  2. #2
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    Quote Originally Posted by topsquark View Post
    You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

    The problem is to put
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}
    in rectangular form.

    I'm kinda stuck on this one.

    Before I mention what I've tried, let me say what the book's answer for this one is:
    (2 - \sqrt{3})^{12}

    My plan of attack was to write \sqrt{3 - i} in rectangular form, giving me
    \sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )
    (By the way, this approach doesn't use that there are actually two values for \sqrt{3 - i}. I'm not sure which one to pick, if it even makes a difference.)

    Now that I've got that, then I can rewrite the original problem:
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =  \left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}

    Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

    And it certainly isn't the nice form (2 - \sqrt{3})^{12}.

    Any thoughts as to how to approach this? Thanks!

    -Dan
    My first thought was to see what \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{2} was - mind you, knowing the answer helped suggest that thought.

    Imagine my disappointment when I got chicken vomit as the answer. Then I decided to put \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} in my TI-89 ......

    Imagine my surprise when I got an answer with a non-zero imaginary part .....

    I'd suggest the question has a typo in it .....
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  3. #3
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    Quote Originally Posted by topsquark View Post
    You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

    The problem is to put
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}
    in rectangular form.

    I'm kinda stuck on this one.

    Before I mention what I've tried, let me say what the book's answer for this one is:
    (2 - \sqrt{3})^{12}

    My plan of attack was to write \sqrt{3 - i} in rectangular form, giving me
    \sqrt{3 - i} = 10^{1/4}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) + i \cdot 10^{1/4}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right )
    (By the way, this approach doesn't use that there are actually two values for \sqrt{3 - i}. I'm not sure which one to pick, if it even makes a difference.)

    Now that I've got that, then I can rewrite the original problem:
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24} =  \left [ 1 - \frac{10^{1/4}}{2}~cos \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) - i \cdot \frac{10^{1/4}}{2}~sin \left ( \frac{1}{2} tan^{-1} \left ( -\frac{1}{3} \right ) \right ) \right ] ^{24}

    Now, believe it or not I can actually do something with this, but it's too complicated for me to bother with the LaTeX. Suffice it to say I can put the base in trigonometric form and use De Moivre's theorem to finish up the problem. But the answer is inordinately messy and I'm frankly not sure I got all of it right with all the picky details to keep track of.

    And it certainly isn't the nice form (2 - \sqrt{3})^{12}.

    Any thoughts as to how to approach this? Thanks!

    -Dan
    A thought for checking the consistency of question and answer ...... a 24th root of the answer should be 1 - \frac{\sqrt{3 - i}}{2}, right?
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    Quote Originally Posted by topsquark View Post
    You know, I'm very glad I'm going through this little book. I thought I knew everything about doing these, but I had never considered a problem like this one.

    The problem is to put
    \left ( 1 - \frac{\sqrt{3 - i}}{2} \right ) ^{24}
    in rectangular form.

    I'm kinda stuck on this one.

    Before I mention what I've tried, let me say what the book's answer for this one is:
    (2 - \sqrt{3})^{12}
    \left | 1 - \frac{\sqrt{3 - i}}{2} \right |\approx 0.187755

    0.187755^{24} \approx 3.68317 \times 10^{-18}

    2 - \sqrt{3} \approx 0.267949

    0.267949^{12} \approx 1.36971\times 10^{-7}

    So I agree with mr fantastic, you have a typo.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    \left | 1 - \frac{\sqrt{3 - i}}{2} \right |\approx 0.187755

    0.187755^{24} \approx 3.68317 \times 10^{-18}

    2 - \sqrt{3} \approx 0.267949

    0.267949^{12} \approx 1.36971\times 10^{-7}

    So I agree with mr fantastic, you have a typo.

    RonL
    Nice back-of-the-envelope confirmation, CaptainBlack. So the question now is:

    What's the typo, that is, what was the intended question? Working back from the answer would probably get it .....
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    Quote Originally Posted by mr fantastic View Post
    So the question now is:

    What's the typo, that is, what was the intended question?
    As a first guess, \left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}. That has the correct absolute value, at any rate.
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    Quote Originally Posted by Opalg View Post
    As a first guess, \left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}. That has the correct absolute value, at any rate.
    Great get, Opalq ol' bean! That is indeed the answer to what question gives the answer.
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    Quote Originally Posted by Opalg View Post
    As a first guess, \left ( 1 - \frac{\sqrt3 - i}{2} \right ) ^{24}. That has the correct absolute value, at any rate.
    There we go! And that makes the problem MUCH simpler. (I'll double check the problem, but I think I typed it in here right.) The trouble with my seeing that on my own is that I was doing problems like \sqrt[8]{3 - 4i} as well, so it could very well have been the problem as I wrote it. But I can do the revised problem without any undue difficulty.

    -Dan
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    Forum Admin topsquark's Avatar
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    Thank you all for your help!

    -Dan
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