# Imaginary Unit

## Summary

The imaginary unit $i$ is a number that is not a real number. The real number system $\mathbb{R}$ is extended to the complex number system $\mathbb{C}$ taking $i$ into account:

$\mathbb{C} = \{ a + bi | a,b \in \mathbb{R} \}.$

The definition of $i$ is quite ambiguous. $i$ is "a" solution to the quadratic equation $x^2 = -1 .$ No square of real number can be negative, so this equation does not have a real solution. Anyway, we have $i^2 = -1 ,$ following the definition of $i.$

Many properties on $\mathbb{R}$ are applied to $\mathbb{C}.$ For example, $(-1)a = -a,$ for all $a \in \mathbb{C} ,$ and $(ab)^2 = a^2 b^2 ,$ for all $a, b \in \mathbb{C} .$ Then we have

$(- i)^2 =((-1)i)^2 = (-1)^2 i^2 = 1 \times (-1) = -1 .$

What's weird is that both equations $i^2 = -1$ and $(-i)^2 = -1$ hold. Which one is the true imaginary unit among $i$ and $-i ?$ Anyway, people have decided to define $i$ as "the" imaginary unit. Actually it is not a problem to define $-i$ as the imaginary unit. $i$ is literally imaginary and does not exist in the real world. $i$ is an abstract concept, and the complex number systems with $i$ and $-i$ as their imaginary units have the same properties.

Addition and multiplication rules on $\mathbb{C}$ can be drawn from associativity, distributivity, and commutativity, which still hold for $\mathbb{C}.$ For all $a=p + qi$ and $b=r + si,$ where $p, q, r, s \in \mathbb{R},$

$\begin{aligned} a + b &= (p + qi) + (r + si) \\ &= (p + r) + (qi + si)\\ &= (p + r) + (q+s)i, \end{aligned}$

and

$\begin{aligned} ab &= (p + qi)(r + si) \\ &= pq + psi + qri + qsi^2 \\ &= pq + (ps+qr)i + qs(-1) \\ &= pq - qs + (ps+qr)i . \end{aligned}$

For a complex number $a=p + qi,$ where $p, q \in \mathbb{R},$ $p$ is called the real part of $a$, denoted by $\text{Re}(a),$ and $q$ is called the imaginary part of $a,$ denoted by $\text{Im}(a).$ A pure imaginary number is a complex number having its real part zero.

## If $x^2 = -4 ,$ what is $x?$

From $2^2 = 4$ and $(\pm i)^2 = - 1,$ we have

$(2(\pm i))^2 = 2^2 (\pm i)^2 = 4 \times (-1) = -4,$

so the answer is $x = \pm 2i .$ $_\square$

## What is $(2+3i) + (5+8i) ?$

We can add two complex numbers by adding real parts and imaginary parts separately: $\begin{aligned} (2+3i) + (5+8i) =& (2+5) + (3+8)i \\ =& 7 + 11i . \ _\square \end{aligned}$

## What is $(4+3i) \times (2-7i) ?$

We can multiply two complex numbers by the distributive law: $\begin{aligned} (4+3i) \times (2-7i) =& (4 \times 2 + 3 \times 7) + (-4 \times 7 + 3 \times 2)i \\ =& 29 - 22i . \ _\square \end{aligned}$

## What is the additive inverse of $-7+9i ?$

The sum of $x+yi$ and $-7+9i$ is

$(x+yi) + (-7 + 9i) = (x-7) + (y+9)i .$

If $x-7=0$ and $y+9=0,$ then the sum equals the additive identity, 0. Therefore, the additive inverse of $-7+9i$ is $x+yi = 7-9i.$ $_\square$

## What is the multiplicative inverse of $3-4i ?$

The product of $x+yi$ and $3-4i$ is

$(x+yi) (3-4i) = (3x+4y) + (-4x+3y)i .$

If $3x+4y=1$ and $-4x+3y=0,$ then the sum equals the multiplicative identity, 1.

Substituting $y = \frac{4}{3}x$ into $3x+4y=1$ gives

$\begin{aligned} 3x + 4\left(\frac{4}{3}x\right) =& 1 \\ \frac{25}{3}x =& 1 \\ x =& \frac{3}{25}. \end{aligned}$

Substituting $x = \frac{3}{25}$ into $y = \frac{4}{3}x$ gives $y = \frac{4}{25} .$

Therefore, the multiplicative inverse of $3-4i$ is $x+yi = \frac{3}{25}\ + \frac{4}{25}i .$ $_\square$

## Find the condition on $x$ and $y$ for $(5-6i)(x+yi)$ to be pure imaginary.

The product of $5-6i$ and $x+yi$ is

$(5-6i)(x+yi) = (5x+6y) + (-6x+5y)i .$

If $5x+6y=0,$ then the real part equals zero, so the product becomes pure imaginary.