(x+1-i)(x+1+i)

Jasonbradley

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Jun 16, 2024

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Expanding the Expression (x+1-i)(x+1+i)

This expression involves complex numbers and can be expanded using the distributive property or by recognizing a special pattern. Let's explore both methods:

Using the Distributive Property

We can expand the expression by multiplying each term in the first set of parentheses by each term in the second set of parentheses. This gives us:

(x + 1 - i)(x + 1 + i) = 
x(x + 1 + i) + 1(x + 1 + i) - i(x + 1 + i)

Now, we distribute each term:

= x^2 + x + xi + x + 1 + i - ix - i - i^2 

Simplifying the expression by combining like terms and noting that i^2 = -1:

= x^2 + 2x + 1 + i^2
= x^2 + 2x + 1 - 1
= x^2 + 2x

Therefore, the expanded form of (x+1-i)(x+1+i) is x^2 + 2x.

Recognizing a Special Pattern

The expression (x+1-i)(x+1+i) resembles the difference of squares pattern: (a + b)(a - b) = a^2 - b^2.

In this case, let:

• a = x + 1
• b = i

Applying the difference of squares pattern:

(x + 1 - i)(x + 1 + i) = (x + 1)^2 - i^2

Expanding and substituting i^2 = -1:

= x^2 + 2x + 1 - (-1)
= x^2 + 2x + 1 + 1
= x^2 + 2x + 2 

Again, we get x^2 + 2x as the simplified result.

Conclusion

Both methods lead to the same result: x^2 + 2x. The difference of squares pattern provides a more concise approach for this particular expression.