1

$\textbf{Question:}$ Am I using the definition of $\sqrt{-1}$ (see below) correctly here or not by giving four cases for $\sqrt{-1}\cdot \sqrt{-1}$?

Also, if I am using this definition correctly, is it true that $(ab)^c=a^cb^c$ for any $a, b, c \in \mathbb{C}$?

Below is the following definition I use to define $\sqrt{-1}$ over $\mathbb{C}$. I was told to use this definition previously on this site.

$\textbf{Definition:}$ $\sqrt{-1}=\pm i$.

So, the other day someone wrote $\sqrt{-1}\cdot \sqrt{-1}=i\cdot i =-1$ and said all other solutions were not possible.

However, I feel that this can be broken down into $4$ cases all of which are acceptable:

Case 1: $\sqrt{-1}\sqrt{-1}=i\cdot i=-1$.

Case 2: $\sqrt{-1}\sqrt{-1}=(-i)(-i)=-1$.

Case 3: $\sqrt{-1}\sqrt{-1}=i(-i)=-i^2=1$.

Case 4: $\sqrt{-1}\sqrt{-1}=(-i)i=-i^2=1$.

So, are there four cases (2 possibilities) or not here?

Community
  • 1
W. G.
  • 1,766

2 Answers2

4

Every number $z\in\mathbb{C}\setminus\{0\}$ has two square roots. So, unless you specify which square root you have in mind, you should not use the expression $\sqrt z$. On the other hand, if you specify which square root you have in mind for each $z\in\mathbb{C}\setminus\{0\}$, then either $\sqrt{-1}=i$ or $\sqrt{-1}=-i$, but not both.

And, in $\mathbb C$, the equality $(ab)^c=a^cb^c$ is meaningless, unless you define $a^b$ for each $a,b\in\mathbb C$. But that will not be easy.

José Carlos Santos
  • 427,504
  • It's the word EITHER that I was forgetting. Thank you! – W. G. Mar 30 '19 at 15:27
  • Actually, I should be able to alternate how I use $\sqrt{-1}$ every time I see it. Remember, how you just said $(ab)^c=a^c\cdot b^c$ is "meaningless"? For instance, $(1+2i)^\frac{1}{3}\cdot (1+5i)^\frac{1}{5}$ would have 3 different possibilities for the $(1+2i)$ term and 5 different possibilities for the $(1+5i)$ term when multiplying these together. For the same reason, this question is "meaningless" and unless specified I could switch up which root I want. Do you agree with that logic? – W. G. Mar 31 '19 at 14:47
  • 1
    That makes sense, but even so I would avoid using $\sqrt z$ unless I was perfectly clear about which square root I am working with. – José Carlos Santos Mar 31 '19 at 14:49
1

Writing $\sqrt{-1}$ is ambiguous because $-1$ has two square roots (namely, $i$ and $-i$).

When we take the square root of a positive real number (e.g., $\sqrt{16}$) we assume by convention that we take the positive square root, i.e., $\sqrt{16}=4$. If we wanted to refer to the other root, we would have to write $-\sqrt{16}=-4$.

Your four cases are akin to saying $\sqrt{16}=\pm 4$, and so:

  1. $\sqrt{16}\sqrt{16}=4\cdot 4 = 16$
  2. $\sqrt{16}\sqrt{16}=(-4)\cdot (-4) = 16$
  3. $\sqrt{16}\sqrt{16}=4\cdot (-4) = -16$
  4. $\sqrt{16}\sqrt{16} =( -4 )\cdot 4 = -16$.

The only line that is correct is line 1 because by convention $\sqrt{16}=4$.

Similarly, when you write $\sqrt{-1}$ you need to specify which root you are referring to. By convention mathematicians write $i=\sqrt{-1}$, even though it is also true that $(-i)^2=-1$. So again, only line 1 is correct.

This post explains why $(ab)^c = a^cb^c$ does not always hold for complex numbers, but it assumes some knowledge of complex analysis (which I'm afraid is necessary to truly understand what's going on).

kccu
  • 20,808
  • 1
  • 22
  • 41