## geerky42 one year ago -1 = 1

1. geerky42

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

$$cis(x) = \cos x+i\sin x$$

3. anonymous

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

I think that we can write this: $\Large \sqrt 1 = \pm 1$

5. Michele_Laino

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

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

So $$\sqrt1$$ is actually $$\pm1$$... My whole life is a lie.

8. anonymous

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

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

Maybe in this content, square root is defined to return the principal root as output?

11. anonymous

yes in your example the left side is the principal square root. the right side is all the roots

12. anonymous

nice problem :)

13. IrishBoy123

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

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

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