The negative square root of $-1$ as the value of $i$

Question

I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i = (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i = -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i = -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

Answer 1

... and we usually take the positive square root of −1 as the value of "$i$".

There's no such thing as a positive or negative complex number. By treating $\mathbb{R}$ as a subset of $\mathbb{C}$ you can call some complex numbers positive or negative, but only those which actually are real numbers. There's no way to define a total order on $\mathbb{C}$ which would behave like the usual order on $\mathbb{R}$ does.

What happens is that one simply picks any root of $-1$, i.e. any solution of $x^2 + 1 = 0$, and calls it $i$. There's then always a second solution, and that solution is $-i$. You can't even say which solution you picked for $i$ - the two solutions of $x^2 + 1=0$ are algebraically indistinguishable, i.e. you cannot tell them apart with algebraic means.

Imagine a friend hands you a bag containing two balls, of equal size and material. You can pick one arbitrarily and call it "ball 1", and take a pen and mark it with a big "1". The other is then "balls 2", and gets marked with a big "2". Now, imagine you had picked the other ball. Would you end up in a different situation? No! You'd still have two balls, one marked "1" and one marked "2", and indistinguishable otherwise. Now, after you've marked the balls, they are of course different. You can now for example put both back into the bag, let your friend pick one, and ask "Which ball have you picked?". But that only works after you marked them!

Answer 2

You are correct that $\sqrt{-1} = -i$, but you are not correct when you say that this fact is neglected--it's used all the time. E.g. the quadratic formula:

$$x^2+x+1=0$$

$$x=\frac{1}{2(1)}+ \frac{\sqrt{1^2-4(1)(1)}}{2(1)}$$

$$x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$$

The plus-or-minus is there because $\sqrt{-3}$ has two roots: $i\sqrt{3}$ and $-i\sqrt{3}$.

Answer 3

The square roots with the properties we know are used only for Positive Real numbers.

We say that $i$ is the square root of $-1$ but this is a convention. You cannot perform operations with the usual properties of radicals if you are dealing with complex numbers, rather than positive reals.

It is a fact that, $-1$ has two complex square roots. We just define one of them to be $i$.

You have to regard the expression $i=\sqrt {-1}$ just as a symbol, and not do operations.

Counterexample: $$\dfrac{-1}{1}=\dfrac{1}{-1}\Rightarrow$$ $$\sqrt{\dfrac{-1}{1}}=\sqrt{\dfrac{1}{-1}}\Rightarrow $$ $$\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}\Rightarrow$$ $$\dfrac{i}{1}=\dfrac{1}{i}\Rightarrow$$ $$i^2=1\text { ,a contradiction}$$