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I came across a really interesting notion on Quora...



To cut it short... this guy says that the 2nd root of -1 is equal to plus or minus i

This is because (i*i)=-1 AND (-i*-i)=(-1*-1*i*i)=-1


Looks right right?


He also says that i squared is equal to plus or minus 1.

-1 because i^2 = -1 like normal


1 because i^2 = i^(4/2) = (i^4)^0.5 = (i^(2+2))^0.5=(i^2+i^2)^0.5=(-1*-1)^0.5=1^0.5=1


That math SEEMS right...intill you think back to those memories of sad sad algebra...



Taking just the -3i^2... you get -3*-1=3 OR -3*1=-3 ...Continuing...

=21+3i OR 15+3i

so which one is it? could it be both?

Is the whole idea stupid... all the math looks reasonable...

 Feb 6, 2018

Sqrt(-1) =i By definition and NOT -i

All you have to remember are these 5 rules:

sqrt(-1) =i

i^2 = -1

i^3 = -i

i^4 = 1

i^5 = i

i^6 = - 1. Then it repeats this pattern from i^2 to i^5 of -1, -i, 1, i......and so on ad infinitum.

 Feb 6, 2018

Understandably, square roots cause a lot of confusion. This is one prime example of this occurring. To be clear, by definition, \(i=\sqrt{-1}\). When you calculate the square root of a number, it is implied that the output is the principal root or the positive result. 


For example, 

\(\sqrt{4}=2\\ \sqrt{9}=3\)


However, if you place a negative sign before the square root, then it is implied that the output should be negative. 


\(-\sqrt{4}=-2\\ -\sqrt{9}=-3\)


If a plus-minus sign precedes a square root, then both answers are desired:


\(\pm\sqrt{4}=\pm2\\ \pm\sqrt{9}=\pm3\)


This confuses many people, and it leads to misconceptions. Since \(\sqrt{-1}\) implies the principal root, the only answer is \(i\).


Examine the false proof that your "guy" provided, which seems to prove that i equals plus or minus 1 to the untrained eye.


\(i^2\\ i^\frac{4}{2}\\ \left(i^4\right)^{0.5}\\ i^{(2+2)^{0.5}}\\ \textcolor{red}{\left(i^2*i^2\right)^{0.5}}\)


The step highlighted in red in an invalid step. It ignores the part about exponent towers. In line 4, you have to evaluate the innermost exponent first to abide by order of operations. In line 5, there are now parentheses around the \(i^2*i^2\), which changes the overall value of the expression.  If this is unclear for you, maybe I will try a simpler example. 


Consider the following two expressions \(2^{3^2}\text{ and }\left(2^3\right)^2\)? The expressions here have a subtle difference, but the output is different. You compute these differently, too. This is another area where misconceptions are prevalent.


\(2^{\textcolor{red}{3^2}}=2^9=512\\ \textcolor{red}{\left(2^3\right)}^2=8^2=64\)


Notice that in the second expression that the parentheses take precedence over any other operation, so you must do that first. Now, let's relate that to the false proof.


\(i^{(2+2)^{0.5}}=\textcolor{red}{i^{4^{0.5}}}=i^2=-1\\ (i^2*i^2)^{0.5}=(i^{2+2})^{0.5}=\textcolor{red}{\left(i^4\right)^{0.5}}\)


Do you notice the discrepancy now? You can now see that the values are different. Since I have cleared up that confusion, I will now end the debate about which complex number the product simplifies to. I'll jump right to the moment of the confusion, which occurs at \(18+3i-3i^2\)


\(18+3i-3i^2\)By definition, \(i=\sqrt{-1}\), so squaring both sides would mean that \(i^2=-1\).
\(18+3i-3*-1\)Now, simplify as much as possible. 
\(18+3i+3\)Combine the like terms. 
\(21+3i\)This is completely simplified, and nothing more can be done here.
 Feb 6, 2018
edited by TheXSquaredFactor  Feb 6, 2018

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