Question

What is the value of \((2+8 i)(1-4 i)-(3-2 i)(6+4 i)\) ? (Note: \(i=\sqrt{-1}\) ) A) 8 B) 26 C) 34 D) 50

Short answer

The value of the expression is 8 (Option A).

Step-by-step solution

Multiply the first two complex numbers

To multiply the first two complex numbers, \((2+8i)(1-4i)\), we apply the distributive property (also known as the FOIL method for binomials): \((2+8i)(1-4i) = 2(1) + 2(-4i) + 8i(1) + 8i(-4i)\)

Simplify the first part

Now we simplify the expression we got in the previous step: \(2(1) + 2(-4i) + 8i(1) + 8i(-4i) = 2 - 8i + 8i - 32i^2\) We know that \(i^2 = -1\), so we substitute this into our expression: \(2 - 8i + 8i - 32(-1) = 2 + 32 = 34\)

Multiply the second two complex numbers

To multiply the second two complex numbers, \((3-2i)(6+4i)\), we again apply the distributive property: \((3-2i)(6+4i) = 3(6) + 3(4i) - 2i(6) - 2i(4i)\)

Simplify the second part

Now we simplify the expression we got in the previous step: \(3(6) + 3(4i) - 2i(6) - 2i(4i) = 18 + 12i - 12i - 8i^2\) Again, we substitute \(i^2\) with \(-1\): \(18 + 12i - 12i - 8(-1) = 18 + 8 = 26\)

Subtract the second part from the first part

Now that we have simplified both parts, we can subtract the second part from the first part: \(34 - 26 = 8\) So the value of the given expression is 8, which corresponds to option A.

Key concepts

Distributive Property

The distributive property is a fundamental principle in algebra. It states that for any numbers or expressions \(a, b,\) and \(c\), the expression \(a(b + c)\) can be expanded to \(ab + ac\). This property also works perfectly with subtraction: \(a(b - c) = ab - ac\).
This becomes especially useful with complex numbers, where each component of one expression must be multiplied with every component of the other, often referred to as the "FOIL" method when dealing with binomials.

When we apply the distributive property to complex numbers like \((2+8i)(1-4i)\), we do:

• First: multiply the first terms \(2 \cdot 1\).
• Outer: multiply the outer terms \(2 \cdot -4i\).
• Inner: multiply the inner terms \(8i \cdot 1\).
• Last: multiply the last terms \(8i \cdot -4i\).

This results in: \(2 - 8i + 8i - 32i^2\). It’s a straightforward application of multiplying each part through, keeping track of signs and terms.

Complex Number Multiplication

Complex numbers are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, satisfying \(i^2 = -1\). When multiplying complex numbers, each part of the first complex number is multiplied with each part of the second complex number, creating four products, as in our example.

Using \((3-2i)(6+4i)\) as an illustration:

• First, multiply the real parts: \(3 \cdot 6 = 18\).
• Outer, multiply the real part of the first with the imaginary part of the second: \(3 \cdot 4i = 12i\).
• Inner, multiply the imaginary part of the first with the real part of the second: \(-2i \cdot 6 = -12i\).
• Last, multiply the imaginary parts: \(-2i \cdot 4i = -8i^2\).

Finally, remember \(i^2 = -1\) so \(-8i^2\) simplifies to \(8\). Combining these, you get \(18 + 12i - 12i + 8\), which simplifies to \(26\). Hubbard's rule applies, tracking all parts carefully.

Simplification of Complex Expressions

Simplification is crucial after multiplying complex numbers to put them into a standard form \(a + bi\). This involves combining like terms and applying the property \(i^2 = -1\) to simplify products of the imaginary unit.

In the example \(2 - 8i + 8i - 32i^2\), observe:

• The terms \(-8i + 8i\) cancel each other out as they sum to zero.
• Recognize \(-32i^2\) as \(-32(-1) = 32\).
• Thus, combining the real numbers gives \(2 + 32 = 34\).

Similarly, from \(18 + 12i - 12i - 8i^2\), follow:

• The \(12i - 12i\) parts cancel out.
• Change \(-8i^2\) into \(8\).
• Add real number parts to result in \(18 + 8 = 26\).

Finally, to solve, subtract these results: \(34 - 26 = 8\). Simplification turns complex, messy expressions into neatly arranged real and imaginary components.