Problem 46 - Algebra and Trigonometry Real Mathematics, Real People

Question

Perform the operation and write the result in standard form. $$(1-2 i)^{2}-(1+2 i)^{2}$$

Step-by-Step Solution

Answer

So, the result of the given operation is \(-8\).

Step 1: Squaring the complex numbers

Square each of the complex numbers separately.\nFinding the square of \((1-2i)\) and \((1+2i)\). Use the formula, \((a-b)^2 = a^2 - 2ab + b^2\) and \((a+b)^2 = a^2 + 2ab + b^2\).\nFor \((1-2i)^2\) it's: \((1 - 2*1*(-2i) - 2^2*i^2)\), which simplifies to \(1 + 4i -4 \).\nAnd for \((1 + 2i)^2\) it's: \((1 +2*1*2i + 2^2*i^2)\), which simplifies to \(1 + 4i +4 \).

Step 2: Subtraction of the squared complex numbers

Subtract \((1+2i)^2\) from \((1-2i)^2\).\nThis gives \((1 + 4i -4) - (1 + 4i +4) = 1 + 4i - 4 - 1 - 4i - 4\).

Step 3: Simplifying the Expression

The above expression then simplifies to \(-8\).

Key Concepts

Squaring Complex Numbers

To understand squaring complex numbers, we first need to know the standard form of a complex number, which is written as \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property that \( i^2 = -1 \). Squaring a complex number essentially means multiplying the complex number by itself.

When we square \( (1 - 2i) \), we use the distributive property of multiplication over addition to expand the expression \( (a - bi)^2 \) into \( a^2 - 2abi + (bi)^2 \). Since \( i^2 = -1 \), the term \( (bi)^2 \) becomes \( -b^2 \). In this case, \( -2^2 \) gives us \( -4 \), thus simplifying to \( 1 + 4i - 4 \).
Similarly, squaring \( (1 + 2i) \) results in \( 1 + 4i + 4 \), showing how the signs within the binomial affect the outcome. This operation is vital in complex numbers algebra as it sets the foundation for more complex operations.

Complex Number Subtraction

Complex number subtraction is performed by separately subtracting the real and imaginary components of the involved complex numbers. If we have two complex numbers in standard form, \( a + bi \) and \( c + di \), their subtraction is given by \( (a + bi) - (c + di) = (a - c) + (b - d)i \).

In our original equation, after squaring the complex numbers, we had \( (1 + 4i - 4) \) and \( (1 + 4i + 4) \). To subtract these, we subtract the real parts, \( 1 - 1 = 0 \), and the imaginary parts, \( 4i - 4i = 0i \), and also the constants, \( -4 - 4 = -8 \). The imaginary parts cancel each other out, leaving us with only the real part, \( -8 \), which is the result in standard form. This process exemplifies basic arithmetic operations within the realm of complex numbers algebra.

Complex Numbers Algebra

Complex numbers algebra encompasses operations like addition, subtraction, multiplication, and division, along with more advanced operations like squaring, finding roots, and exponentiation. The core principle is to treat \( i \) as a variable that satisfies \( i^2 = -1 \), and then manipulate the expressions following the established rules of algebra.

The role of algebra in complex numbers becomes evident when simplifying expressions, solving quadratic equations, and in our situation, performing operations such as squaring and adding or subtracting results. In the subtracting step of the example, we applied algebraic principles to combine like terms. Common misunderstandings sometimes arise if the imaginary unit \( i \) is not handled correctly, therefore, remembering that \( i^2 = -1 \) is crucial. Algebra within the complex number system essentially extends the concepts we know from real numbers into a new dimension, giving us tools to solve problems previously not solvable with just real numbers alone.