-1

I am not sure about this:

$$\frac{1}{\sqrt{-1}}=\frac1i=-i $$

or

$$\frac{1}{\sqrt{-1}}=\sqrt{-1}=i $$

?

I would say the second but Wolfram Mathematica says

$$\frac{1}{\sqrt{-1}}=\frac1i=-i $$

What is the convention?

Tianlalu
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Mencia
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7 Answers7

2

Let $i=\sqrt {-1}$.

From $i\cdot i=-1$ divide both sides by $i$ to obtain $i=-\frac 1i$

You need to make a consistent choice of sign, because there are two possible square roots $i$ and $-i$. To avoid errors, use a notation which enforces this.

Mark Bennet
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2

This question strikingly shows what calamity is the notation $\sqrt{-1}$. There is no such mathematical object, and the notation $\sqrt{\phantom{a}}$ should be used exclusively for its original aim: denote the positive square root of a positive real number. In all other contexts, we cannot distinguish between the two square roots of a complex number

Bernard
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  • According to you, what should Wolfram Mathematica do when there is something like $1/ \sqrt{-f^2}$ for real $f$ ? – Mencia Jan 14 '17 at 10:34
  • I really don't know, as this depends very much on the context: in Number theory, one often meets things like $\mathbf{Q}(\sqrt{-3})$, but there is no ambiguity,as it denotes any element satisfying he equation $x^2+3=0$, and it is more of a symbolic, not to say metaphoric, notation. I suppose Mathematica (which I don't use) knows this usage). – Bernard Jan 14 '17 at 10:46
  • I think that Mathematica's algorithm when dealing with fractional roots (say, $1/n$) of negative numbers is to take the fractional root of the absolute value and multiply by the $2n$th root of unity with smallest positive argument. This applies whether $n$ is even or odd. – rogerl Jan 30 '19 at 13:31
1

Note that the powers of $i$ are cyclic.

$$\color{red}{i^0=1}$$ $$\color{blue}{i^1=i}$$ $$\color{green}{i^2=-1}$$ $$\color{orange}{i^3=-i}$$ $$\color{red}{i^4=1}$$ $$\color{blue}{i^5=i}$$ $$...$$

This pattern also applies for negative powers.

$$\color{red}{i^{-4}=1}$$ $$\color{blue}{i^{-3}=i}$$ $$\color{green}{i^{-2}=-1}$$ $$\color{orange}{i^{-1}=-i}$$

Therefore, it is easy to deduce that the result of $i^{2017}=i$ is for example, as well as $\frac{1}{i}=-i$ just by dividing the power by $4$ and looking at the remainder obtained.

projectilemotion
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1

The typographical picture $\sqrt{-1}$ does not define a mathematical entity, but is the formulation of a problem. Therefore it makes no sense to write things like ${1\over\sqrt{-1}}$ or even $e^{\sqrt{-1}\,\phi}$.

When we write "Let $i=\sqrt{-1}$" this is a colloquial expression of the agreement that henceforth $i$ is a constant satisfying $i^2=-1$.

In any case one has ${1\over i}=-i$. This is not a convention but a stringent consequence from $i^2=-1$ and the field axioms valid in the "complex environment".

Christian Blatter
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  • Thank you for your answer. When I asked about the convention, I was not referring to 1/i=-i. I was referring to 1/Sqrt[-1]=i vs 1/Sqrt[-1]=-i. – Mencia Jan 14 '17 at 09:48
  • According to you, what should Wolfram Mathematica do when there is something like $1/ \sqrt{- f^2}$ for real $f$ ? – Mencia Jan 14 '17 at 09:58
1

Whenever we have some negative term in square root. Its not good practice to solve it like you are solving in method 2.

You have to replace -1 into $i^2$.

Edit -

Product Rule (extended) -

$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

where a ≥ 0, b ≥ 0 Or a ≥ 0, b < 0

But NOT a < 0, b < 0

So $\sqrt{-1} \cdot \sqrt{-1} \ne 1$

Kanwaljit Singh
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0

Addition to what other said, we can say for $i=0+i=(0,1)\in\mathbb{C}$: $$\dfrac{1}{\sqrt{-1}}=\frac{1}{0+i}=\frac{1}{0+i}\times\frac{0-i}{0-i}=\frac{-i}{(-1,0)}{\color{red}\equiv}\frac{-i}{-1}=i$$

Nosrati
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0

$\mathbb{C}$ is a field, so exist $a , b \in \mathbb{R}$ such that $\frac{1}{i} = a + i b$. We will show that $a = 0$ and $b = - 1$. We have that $$ \frac{1}{i} = a + i b \Longleftrightarrow 1 = i (a + i b) = i a + i^2 b = - b + i a\mbox{,} $$ so, taking real and imaginary parts, $$ Re (a i - b) = Re 1 \Longleftrightarrow - b = 1 \mbox{ and } Im (a i - b) = Im 1 \Longleftrightarrow a = 0\mbox{,} $$ because of this $a = 0$ and $b = - 1$.

joseabp91
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