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We have the definition that $\imath=\sqrt{-1}$ correct. Then does it follow that $$ \sqrt{1-x}=\imath\sqrt{x-1} $$ It seems to me that this is obvious from the definition and products of square roots since we have $$ \imath\sqrt{x-1}=\sqrt{-1}\sqrt{x-1}=\sqrt{-1(x-1)}=\sqrt{1-x} $$ however can't seem to find this plainly written anywhere. Is this indeed always correct?

I have just randomly chosen the argument of the square root function to be $1-x$ or $x-1$ here. However, does the actual value of this (say being positive/negative or real/complex) change anything or is this property always true?

EDIT: Assuming that $a,b$ are real numbers and that $D=\sqrt{\frac{a^2}{b^2}-1}$ is a positive real number, then is it true that $$ \sqrt{1-\frac{a^2}{b^2}} = \imath D $$ Because I know $D$ is a real positive number, it seems that I know that the expression on the LHS of the equation above has to be purely imaginary and so it seems there should be no reason why this isn't true.

However, is this just one specific case where it holds but in general it isn't always true?

Further, can someone please explain what is meant by @David K's comment and which of my rationalizations are correct and where the other goes wrong (also in comments).

Further, I agree that the other question linked in the comments is essentially asking the same thing, however, I don't think any of the answers there completely satisfy my concerns.

Mjoseph
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    There are many posts on this site that discuss the paradoxes you see when you try to use the $\sqrt{\phantom{a}}$ sign for anything other than a nonnegative real number. – Ethan Bolker Nov 11 '24 at 16:46
  • Try $x=0$. On the left you have $\sqrt1$. On the right you have $i\sqrt{-1}=i\cdot i$. – David K Nov 11 '24 at 17:02
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    Your definition is not correct. Both $i$ and $-i$ are solutions of the equation $x^2=-1$ ; there is no reason to choose $\sqrt{-1}$ to be $i$ and not $-i$ ; $-i$ is as eligible as $i$. BTW, I even don't know if you define $i$ as some square root of $-1$, or if you define the Square-root function for negative numbers, by $\sqrt{x}= i\sqrt{|x|}$ – Lourrran Nov 11 '24 at 17:18
  • @EthanBolker Thank you for pointing me to that question. I don't quite find any of those answers to be definitive however. I am still confused why $\sqrt{a}\sqrt{b} \neq \sqrt{ab}$ if $a$ and $b$ are complex numbers. – Mjoseph Nov 11 '24 at 18:03
  • @DavidK If $x=0$, then the equation reads $\sqrt{1} = \imath \sqrt{-1} = \sqrt{-1} \sqrt{-1} = \sqrt{1}$ which would then be equivalent. However, I could also reason that $\sqrt{1}=\imath \sqrt{-1} = \imath^2 = -1 $ which isn't true. Now I find that I'm more confused than I was before. – Mjoseph Nov 11 '24 at 18:04
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    Your first mistake is in the first sentence of the post ("we have the definition..."). – Moishe Kohan Nov 11 '24 at 18:37
  • That's not the definition of $i$. Also, the standard notation is $i$, not $\imath$ – jjagmath Nov 11 '24 at 18:41
  • If you say that $\sqrt{-1}\sqrt{-1}=\sqrt1$ then you are saying that $(\sqrt{-1})^2=\sqrt1=1$ which contradicts the notion that $\sqrt{-1}$ is the square root of negative one. It is possible to define $\sqrt{\ }$ as a function on complex numbers, but not if you use it the way you're using it. – David K Nov 11 '24 at 19:12

1 Answers1

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$$\sqrt{1-x}=i\sqrt{x-1} \iff x=1$$

There is a rule that says $\sqrt a\times \sqrt b = \sqrt{ab}$, as long as $a$ and $b$ are non-negative Real numbers, but as per the definition that does not work for $i$ as it is not Real. This is because for complex numbers, each square root has two possible values, and the need to choose a specific branch (or aangle) for consistency leads to discontinuities, breaking the rule.

This answer explains in more detail.

Ilikemath
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  • Depending on how you define $\sqrt z$ where $z$ is complex, it is possible to have $\sqrt{1-x}=i\sqrt{x-1}$ in a lot more places than just at $x=1$. It's just not possible (AFAIK) to make it true on the entire complex plane. – David K Nov 11 '24 at 19:19