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geerky42
 one year ago
1 = 1
geerky42
 one year ago
1 = 1

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geerky42
 one year ago
Best ResponseYou've already chosen the best response.4We 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?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4\(cis(x) = \cos x+i\sin x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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. xaxis". 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.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I think that we can write this: \[\Large \sqrt 1 = \pm 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3furthermore, we can write this: \[\Large \sqrt {{e^{i2\pi }}} = \cos \left( {\pi + k\pi } \right) + i\sin \left( {\pi + k\pi } \right),\quad k = 0,1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its 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

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4So \(\sqrt1\) is actually \(\pm1\)... My whole life is a lie.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4Maybe in this content, square root is defined to return the principal root as output?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes in your example the left side is the principal square root. the right side is all the roots

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0hope you don't mind @jayzdd, i've latexed your work...really helped \(1 = cos(0 + 2\pi \ n) + i \ sin(0 + 2\pi \ n) \\ 1^{1/2} = cos(\frac{0}{2} + \frac{2 \ \pi \ n}{2}) + i \ sin(\frac{0}{2} + \frac{2 \ \pi \ n}{2}) \\ 1^{1/2} = cos ( \pi \ n ) + i \ sin(\pi \ n) , \ \ \ \ n= 0,1 \\ 1^{1/2} = 1, 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, 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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