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Math Help - Find the smallest integer n

  1. #1
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    Find the smallest integer n

    Question: Find the smallest integer n for which \left(\frac{1+i}{1-i} \right)^n = 1
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  2. #2
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    Start by converting to polar coordinates:

    \left(\frac{1+i}{1-i} \right)^n = (\frac{\sqrt 2 \arg 45}{\sqrt 2 \arg -45})^n=\frac{(\sqrt 2)^n \arg (45*n)}{(\sqrt 2)^n \arg (-45*n)}=1 \arg (90*n)

    Now we can see that the solutions are n=...-12,-8,-4,0,4,8,12 ...
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  3. #3
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    I am still unable to understand

    Quote Originally Posted by Kiwi_Dave View Post
    Start by converting to polar coordinates:

    \left(\frac{1+i}{1-i} \right)^n = (\frac{\sqrt 2 \arg 45}{\sqrt 2 \arg -45})^n=\frac{(\sqrt 2)^n \arg (45*n)}{(\sqrt 2)^n \arg (-45*n)}=1 \arg (90*n)

    Now we can see that the solutions are n=...-12,-8,-4,0,4,8,12 ...

    Thanks for the post but i still cannot understand
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  4. #4
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    Zorro

    Can you answer "convert 1+i to polar coordinates". If you can show the notation that you are familiar with then it will be easier to answer your question.

    I would say 1+i=\sqrt 2 \arg {45} but you may have been taught to write it in a different way.
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  5. #5
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    My work ,please let me know what i have done wrongly

    Quote Originally Posted by Kiwi_Dave View Post
    Zorro

    Can you answer "convert 1+i to polar coordinates". If you can show the notation that you are familiar with then it will be easier to answer your question.

    I would say 1+i=\sqrt 2 \arg {45} but you may have been taught to write it in a different way.

    \left( \frac{1 + i}{1 - i} \right)^n = \frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})}

    I am stuck here .....what should i do now??????
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  6. #6
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    \left( \frac{1 + i}{1 - i} \right)^n = \frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})} = \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}{(cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))} = \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))^2}{cos^2(\frac{n \pi}{4}) + sin^2 (\frac{n \pi}{4})} = \frac{(cos(\frac{n \pi}{2}) + i sin (\frac{n \pi}{2}))}{1}

    Now you want this to equal 1 so the complex part must be zero. That is

    sin (\frac{n \pi}{2})=0, this occurs if and only if n is even.

    Now find the EVEN values of n such that:

    cos (\frac{n \pi}{2})=1
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  7. #7
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    smallest number

    Quote Originally Posted by Kiwi_Dave View Post
    \left( \frac{1 + i}{1 - i} \right)^n = \frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})} = \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}{(cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))} = \frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))^2}{cos^2(\frac{n \pi}{4}) + sin^2 (\frac{n \pi}{4})} = \frac{(cos(\frac{n \pi}{2}) + i sin (\frac{n \pi}{2}))}{1}

    Now you want this to equal 1 so the complex part must be zero. That is

    sin (\frac{n \pi}{2})=0, this occurs if and only if n is even.

    Now find the EVEN values of n such that:

    cos (\frac{n \pi}{2})=1


    I beleive the smallest the number would n = 0
    Am i correct ???
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  8. #8
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    Hello, zorro!

    Find the smallest integer n for which: . \left(\frac{1+i}{1-i} \right)^n \:=\: 1

    We have: . \frac{(1+i)^n}{(1-i)^n} \:=\:1

    Multiply by (1-i)^n\!:\quad (1+i)^n \:=\:(1-i)^n


    Multiply by (1 + i)^n\!:\quad (1+i)^n\cdot(1+i)^n \:=\:(1+i)^n(1-i)^n

    . . . . . (1+i)^{2n} \:=\:\bigg[(1+i)(1-i)\bigg]^n \;=\;\left(1 - i + i - i^2\right)^n

    And we have: . (1+i)^{2n} \:=\:2^n


    By inspection: . n \,=\,0

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  9. #9
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    n=-4000 is an integer less than 0 and cos(\frac{-4000 \pi}{2})=1 so -4000 is also a solution. There is no unique smallest solution because for any solution n, n-4 is also a solution.

    If your question actually asked for the "smallest positive integer" then the answer would be 4.
    Last edited by Kiwi_Dave; December 15th 2009 at 11:51 PM. Reason: addition
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