Squaring a Complex Number

To square a complex number \( z = (a, b) \), you can use the following formula: $$ (a, b)^2 = (a^2 - b^2, 2ab) $$ Alternatively, if the number is written in algebraic form as \( z = a + bi \), you can expand the square of the binomial \( (a + bi)^2 \), remembering that the square of the imaginary unit is \( i^2 = -1 \): $$ z^2 = (a + bi)^2 $$ $$ z^2 = a^2 + 2abi + (bi)^2 $$ $$ z^2 = a^2 + 2abi - b^2 $$ As you can see, both approaches lead to the same result.

A Practical Example

Let's take the complex number \( z = (2,3) \) and find its square.

$$ z = (2,3) $$

Method 1: Using the Formula

We apply the formula directly:

$$ z^2 = (a, b)^2 = (a^2 - b^2, 2ab) $$

Substituting \( a = 2 \) and \( b = 3 \):

$$ (2,3)^2 = (2^2 - 3^2, 2 \cdot 2 \cdot 3) $$

$$ (2,3)^2 = (4 - 9, 12) $$

$$ (2,3)^2 = (-5,12) $$

So, the squared complex number is \( (-5,12) \):

$$ z^2 = (-5,12) $$

In algebraic form, this is written as:

$$ z^2 = -5 + 12i $$

Method 2: Expanding the Binomial

First, rewrite the complex number in algebraic form:

$$ z = (2,3) = 2 + 3i $$

Now, square it using the binomial expansion: $$ (a + b)^2 = a^2 + 2ab + b^2 $$

Applying this to our complex number:

$$ z^2 = (2 + 3i)^2 $$

$$ z^2 = 2^2 + 2 \cdot 2 \cdot 3i + (3i)^2 $$

$$ z^2 = 4 + 12i + 9i^2 $$

Since \( i^2 = -1 \), we substitute:

$$ z^2 = 4 + 12i + 9(-1) $$

$$ z^2 = 4 + 12i - 9 $$

$$ z^2 = -5 + 12i $$

As expected, we arrive at the same result.

Proof of the Formula

To prove the formula: $$ (a, b)^2 = (a^2 - b^2, 2ab) $$ we start by expressing the square of a complex number as the product of the number with itself:

$$ (a, b)^2 = (a, b) \cdot (a, b) $$

Using the complex number multiplication rule:

$$ (a, b)^2 = (a, b) \cdot (a, b) $$

$$ (a, b)^2 = (a \cdot a - b \cdot b, a \cdot b + b \cdot a) $$

$$ (a, b)^2 = (a^2 - b^2, ab + ab) $$

$$ (a, b)^2 = (a^2 - b^2, 2ab) $$

And that gives us the formula for squaring a complex number.

Alternative Proof: Another way to derive the formula is by first converting the complex number to algebraic form: $$ (a, b)^2 = (a + bi)^2 $$ Then expanding using the binomial theorem: $$ (a, b)^2 = a^2 + 2a(bi) + (bi)^2 $$ $$ (a, b)^2 = a^2 + 2abi + b^2i^2 $$ Since \( i^2 = -1 \), we substitute: $$ (a, b)^2 = a^2 + 2abi + b^2(-1) $$ $$ (a, b)^2 = a^2 + 2abi - b^2 $$ Thus, we arrive at the same formula, proving it once again.

Squaring a Real Complex Number

The square of a real complex number \( (a,0) \) is simply the square of its real part: $$ (a,0)^2 = (a^2,0) $$

In other words, squaring a real complex number always results in a positive real number \( a^2 \).

Proof

We calculate the square by multiplying the real complex number by itself:

$$ (a,0)^2 = (a,0) \cdot (a,0) $$

Expanding the multiplication:

$$ (a,0)^2 = (a \cdot a - 0 \cdot 0, a \cdot 0 + 0 \cdot a) $$

$$ (a,0)^2 = (a^2,0) $$

So, the result is simply the square of the real number, \( a^2 \).

Example

Let's square the real complex number \( (2,0) \) using the formula:

$$ (2,0)^2 = (2^2,0) $$

$$ (2,0)^2 = (4,0) $$

So, the square of \( (2,0) \) is the real complex number \( (4,0) \).

Verification: To confirm the result, let's compute the product directly: $$ (2,0)^2 = (2,0) \cdot (2,0) $$ $$ (2,0)^2 = (2 \cdot 2 - 0 \cdot 0, 2 \cdot 0 + 0 \cdot 2) $$ $$ (2,0)^2 = (4 - 0, 0 + 0) $$ $$ (2,0)^2 = (4,0) $$ The result is exactly as expected.

Squaring an Imaginary Number

The square of a purely imaginary number \( (0,b) \) follows this rule: $$ (0,b)^2 = (-b^2,0) $$

In simpler terms, squaring an imaginary number always results in a negative real number \( -b^2 \).

Proof

We calculate the square by multiplying the imaginary number by itself:

$$ (0,b)^2 = (0,b) \cdot (0,b) $$

Expanding the multiplication:

$$ (0,b)^2 = (0 \cdot 0 - b \cdot b, 0 \cdot b + b \cdot 0) $$

$$ (0,b)^2 = (0 - b^2, 0 + 0) $$

$$ (0,b)^2 = (-b^2, 0) $$

This confirms the expected result.

Example

Let's square the imaginary number \( (0,2) \) using the formula:

$$ (0,2)^2 = (-2^2,0) $$

$$ (0,2)^2 = (-4,0) $$

So, the square of \( (0,2) \) is the real complex number \( (-4,0) \).

Verification: To confirm the result, let's compute the product directly: $$ (0,2)^2 = (0,2) \cdot (0,2) $$ $$ (0,2)^2 = (0 \cdot 0 - 2 \cdot 2, 0 \cdot 2 + 2 \cdot 0) $$ $$ (0,2)^2 = (0 - 4, 0 + 0) $$ $$ (0,2)^2 = (-4,0) $$ As expected, the result is the same.

Square Roots of Negative Numbers

Squaring imaginary numbers also helps us understand how to compute the square root of a negative real number using complex numbers.

Example

Let's find the square root of the negative real number \(-4\):

$$ x = \sqrt{-4} $$

In the real number system, this operation is impossible. However, in the world of complex numbers, we can solve it.

We begin by rewriting the expression:

$$ x = \sqrt{-1 \cdot 4} $$

Since we know that \(-1\) is the square of the imaginary unit \( i^2 = -1 \), we substitute:

$$ x = \sqrt{i^2 \cdot 4} $$

Using the square root property:

$$ x = i \cdot \sqrt{4} $$

Since the square root of 4 is \( \pm 2 \), we get:

$$ x = i \cdot (\pm 2) $$

$$ x = \pm 2i $$

So, the solutions to \( \sqrt{-4} \) are the complex numbers \( 0 + 2i \) and \( 0 - 2i \), or equivalently, the imaginary numbers \( (0,2) \) and \( (0,-2) \).

$$ x = \sqrt{-4} = 0 \pm 2i $$

And so on.