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Math Help - help solve (i-j)^n=(iz-1)^n

  1. #1
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    help solve (i-j)^n=(iz-1)^n

    Hi,

    I can't solve (i-j)^n=(iz-1)^n nor (z+i)^n + (z-i)^n =0.

    I have tried dividing through and equating with the nth root but it hasn't gotten me anywhere?

    Anoone know what to do??

    Thanks..
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  2. #2
    MHF Contributor red_dog's Avatar
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    (i-z)^n=(iz-1)^n

    Let z=x+yi.

    Apply modulus in both members:
    |i-z|=|iz-1]\Rightarrow|-x+(1-y)i|=|-1-y+xi|\Rightarrow \Rightarrow x^2+1-2y+y^2=1+2y+y^2+x^2\Rightarrow y=0\Rightarrow z=x

    Divide equation by the right member:

    \left(\frac{i-x}{ix+1}\right)^n=1\Rightarrow\frac{i-x}{ix-1}=z_k, where z_k=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}, \ k\in{0,1,....,n-1}

    Find x: x=\frac{i+z_k}{1+iz_k}

    x=\frac{\cos\frac{2k\pi}{n}+i\left(1+\sin\frac{2k\  pi}{n}\right)}{1-\sin\frac{2k\pi}{n}+i\cos\frac{2k\pi}{n}}=

    =\frac{\cos^2\frac{k\pi}{n}-\sin^2\frac{k\pi}{n}+i\left(\cos\frac{k\pi}{n}+\si  n\frac{k\pi}{n}\right)^2}{\left(\cos\frac{k\pi}{n}-\sin\frac{k\pi}{n}\right)^2+i\left(\cos^2\frac{k\p  i}{n}-\sin^2\frac{k\pi}{n}\right)}=

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

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

    Solve the other equation in a similar way.
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  3. #3
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    big thanks
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  4. #4
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    (tears of joy...) big thanks!
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