Imaginary Unit

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 \(p, q, r, s \in \mathbb{R},\)

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

and

\[\begin{align} ab &= (p + qi)(r + si) \\ &= pr + psi + qri + qsi^2 \\ &= pr + (ps+qr)i + qs(-1) \\ &= pr - qs + (ps+qr)i . \end{align} \]

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 \(x?\)

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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 \( (2+3i) + (5+8i) ?\)

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

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

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

What is the additive inverse of \( -7+9i ?\)

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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 \( 3-4i ?\)

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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

\[\begin{align} 3x + 4\left(\frac{4}{3}x\right) =& 1 \\ \frac{25}{3}x =& 1 \\ x =& \frac{3}{25}. \end{align}\]

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.

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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.