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geerky42

  • one year ago

-1 = 1

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  1. geerky42
    • one year ago
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    We know that \(cis(2\pi)=1\), so we have... \[cis(2\pi)=1\quad\Rightarrow\quad\sqrt{cis(2\pi)}=\sqrt1=1\] By De Moivre's formula, we have \(\sqrt{cis(2\pi)} = cis(\pi)=-1\) Therefore \(\boxed{-1=1}\) What did I do wrong here?

  2. geerky42
    • one year ago
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    \(cis(x) = \cos x+i\sin x\)

  3. anonymous
    • one year ago
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    This may be incomplete, but cis is used to indicate direction on an Armand diagram (polar coordinates). So cis(pi) = cos(pi) + isin(pi) = -1 + 0 = direction along the "neg. x-axis". The coefficient of cis specifies the length on the Armand diagram. Hence, 3cis(pi) represents 3 units in the negative x direction. And taking the square root of a direction is meaningless.

  4. Michele_Laino
    • one year ago
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    I think that we can write this: \[\Large \sqrt 1 = \pm 1\]

  5. Michele_Laino
    • one year ago
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    furthermore, we can write this: \[\Large \sqrt {{e^{i2\pi }}} = \cos \left( {\pi + k\pi } \right) + i\sin \left( {\pi + k\pi } \right),\quad k = 0,1\]

  6. anonymous
    • one year ago
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    its more illuminating if you apply Demoivre's as follows 1 = cos(0 + 2pi*n) + i* sin(0 + 2pi* n) 1^(1/2) = cos(0/2 + 2pi*n/2) + i* sin(0/2 + 2pi* n/2) 1 1^(1/2) = cos ( pi* n ) + i * sin(pi *n) , n= 0,1 1^(1/2) = 1, -1 that makes sense because (1)^2 = 1 , (-1)^2 = 1

  7. geerky42
    • one year ago
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    So \(\sqrt1\) is actually \(\pm1\)... My whole life is a lie.

  8. anonymous
    • one year ago
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    The square root, cube root, or nth root of a number can give you n solutions. The square root button as you see on your calculator gives you one solution, that is how the calculator defines square root. IF you use wolfram, the square root of 1 has 2 solutions.

  9. anonymous
    • one year ago
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    It depends on the context. If you are looking at y = sqrt(x), this square root function has one output for each input. So sqrt(1) = 1 , not two outputs 1,-1 But technically there are two numbers which square to make 1

  10. geerky42
    • one year ago
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    Maybe in this content, square root is defined to return the principal root as output?

  11. anonymous
    • one year ago
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    yes in your example the left side is the principal square root. the right side is all the roots

  12. anonymous
    • one year ago
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    nice problem :)

  13. IrishBoy123
    • one year ago
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    hope you don't mind @jayzdd, i've latexed your work...really helped \(1 = cos(0 + 2\pi \ n) + i \ sin(0 + 2\pi \ n) \\ 1^{1/2} = cos(\frac{0}{2} + \frac{2 \ \pi \ n}{2}) + i \ sin(\frac{0}{2} + \frac{2 \ \pi \ n}{2}) \\ 1^{1/2} = cos ( \pi \ n ) + i \ sin(\pi \ n) , \ \ \ \ n= 0,1 \\ 1^{1/2} = 1, -1\)

  14. anonymous
    • one year ago
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    the error is in assuming that \(\sqrt{e^{2\pi i}}=e^{i\pi}\), i.e. assuming the principal square root (if it exists) of \(e^{2k\pi i}=e^{k\pi i}\). this is a misapplication of de Moivre's theorem, which gives solutions to \(z^k=re^{it}\); there can only be one principal root or solution to that equation, though, which is what \(\sqrt[k]{re^{it}}\) gives

  15. anonymous
    • one year ago
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    so, no, \(\sqrt1\) is not ambiguous, it is defined as \(\sqrt1=1\). this does not conflict with the fact that \((-1)^2=1\), since there can be multiple possible roots and yet only one *principal* root. http://en.wikipedia.org/wiki/Branch_cut

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