# A proper definition of $i$, the imaginary unit

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• Refining my knowledge of the imaginary number

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To address your final question, the natural extension of the square root function to the complex numbers is done by writing it in polar form, $z=re^{i\theta}$, with $\theta \in [0,2\pi)$, $\sqrt z=\sqrt re^{i \theta/ 2}$. Then if you have a real positive number $A$, $-A=Ae^{i\pi}$, so $\sqrt {-A}=\sqrt Ae^{i \pi /2}=\sqrt Ai$, so no, this is never a problem with positive numbers.

This naturally shows where the problem with breaking apart two square roots occur in the complex numbers: In order to have $\sqrt {z_1z_2}=\sqrt {z_1} \sqrt {z_2}$, we need $Arg(z_1)+Arg(z_2)<2 \pi$, because otherwise we have a problem with the principal arguments changing.
(That’s an if and only if, btw)

The mathematically correct way is to call $i$ a root of $x^{2}+1$, or equivalently to call it a symbol satisfying $i^{2} = -1$. There is no determinacy, nor does there need to be, because $i$ and $-i$ are indistinguishable algebraically (they are both conjugate roots of the same irreducible polynomial in $\mathbb{R}[x]$). There is no “principal” root of $x^{2}+1$.

You need to be careful when you consider $\sqrt{xy}$, obviously; it is only stated to be equal to $\sqrt{x}\sqrt{y}$ when $x$ and $y$ have real square roots.

Finally, complex numbers: sure, algebraic completion of the reals. Perfectly sound definition. A field isomorphic to $\mathbb{R} \times \mathbb{R}$: makes no sense, this is not a field without adding multiplication rules, and adding those is really not motivated unless you ALREADY have $\mathbb{C}$ in mind, which makes it rather circular.

This a definition of a complex number which can resolve many of the problems which you mentioned:

A complex number is a ordered pair of two real numbers: $(a,b),\,\, a,b\in \mathbb R$, with the following definitions of arithmetical operations:

Complex numbers can be added: $(a,b) + (c,d) = (a+c,b+d)$.
They can be multiplied by a real number: $c(a,b)=(ca,cb)\,\,c\in\mathbb R$.
They can also by multiplied: $(a,b) \cdot (c,d) = (ac-bd,ad+bc)$.

Then let the number (1,0) be denoted by 1 and the number (0,1) be denoted by $i$. 1

It follows from the definition of multiplication that $i\cdot i=(-1,0)=-1$.

With this definition, there is no need for $i$ to be “imaginary” and also it is totally determined that only (0,1) is $i$ and not (0,-1) so it removes the ambiguity of $i=\sqrt{-1}$.

1 Note: Using this, we have $(a,b) = a(1,0) + b(0,1) = a \times 1 + b \times i$ which can be written as $a+bi$.

The set of complex numbers is R² (cartesian product of the set of real numbers), once defined the specific operations:

(x, y)+(x', y') = (x+x', y+y')
(x, y)*(x', y') = (xx'-yy', xy'+yx')


i is only a very useful notation for the complex number (0,1).
And you can check that
i² = (0,1)*(0,1) = (-1, 0) = -1 + 0*i = -1

Complex numbers were introduced to solve second-degree equations in specific cases, in the same way that you would define the set of real numbers (or at least rational numbers) to solve equations as x = 3^(-1) when you only know integers.

Obviously sqrt(-1) is not defined, as the sqrt function is only defined for positive numbers (including 0). So, to answer the edit, sqrt(-A) where A > 0 is not defined. When you have x² = -A you do not solve the equation with sqrt(-A) = i * sqrt(A). You look for numbers such that x² = -A and i * sqrt(A) is one of them (-i * sqrt(A) is the other one).

We define $\Bbb C$ to be $\Bbb R \times \Bbb R$ with the usual multiplication which makes it into a field. You seem to be familiar with this.

Once we have this, we simply define $i$ to be one of the roots of the equation $x^2 + 1 = 0$ in $\Bbb C$. There is no algebraically independent way to choose a “principal root” of this equation, because indeed, there is a field automorphism of $\Bbb C$ characterized by $i \mapsto -i$. Thus, we can choose $i$ to be either root of the equation without changing anything which really matters.

It is not what we do believe is better or what we want to define as $i$.
The definition says that $i$ is a number with the property $i^2=-1$.
Stick with the definition and proceed.No square roots are needed.As you will see,everything you want to prove or solve in complex analysis uses only $i^2=-1$.
This is why definitions are not made in one minute.Because we use the least we can to obtain the more we want.