Multiply each pair of factors. Type the product in the space provided.

$\begin{array}{l}

(6+3 i)(6-3 i)= \\

(4-5 i)(4+5 i)=\square \\

(-3+8 i)(-3-8 i)=

\end{array}$

Verified Answer :

  • Use the identity \((a+bi)(a-bi) = a^2+b^2\) to compute products of complex conjugates.
  • For \((6+3i)(6-3i)\), compute \(6^2 + 3^2 = 36 + 9 = \boxed{45}\).
  • For \((4-5i)(4+5i)\), compute \(4^2 + 5^2 = 16 + 25 = \boxed{41}\).
  • For \((-3+8i)(-3-8i)\), compute \((-3)^2 + 8^2 = 9 + 64 = \boxed{73}\).

Explanation

  1. Understanding Complex Conjugates and the Multiplication Property

The problem asks us to compute the product of three pairs of complex conjugate numbers. Recall that a complex number \(z\) can be written in the form \(z = a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit with \(i^2 = -1\). The complex conjugate of \(z = a + bi\) is \(z^* = a - bi\). The product of a complex number and its conjugate is a real number given by the expression:
\[(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2.\]
We will use this property to compute each product provided.

  1. Computing the First Product

For the first expression, \((6+3i)(6-3i)\), we have \(a = 6\) and \(b = 3\). Applying the formula \(a^2 + b^2\):
\[(6+3i)(6-3i) = 6^2 + 3^2 = 36 + 9 = 45.\]

  1. Computing the Second Product

For the second expression, \((4-5i)(4+5i)\), we note that this is the same as \((4+(-5)i)(4-(-5)i)\) or more simply just identify \(a=4\) and \(b=5\). Applying the formula \(a^2 + b^2\):
\[(4-5i)(4+5i) = 4^2 + 5^2 = 16 + 25 = 41.\]

  1. Computing the Third Product

For the third expression, \((-3+8i)(-3-8i)\), we have \(a = -3\) and \(b = 8\). Applying the formula \(a^2 + b^2\):
\[(-3+8i)(-3-8i) = (-3)^2 + 8^2 = 9 + 64 = 73.\]

  1. Final Results

The products of the given pairs are 45, 41, and 73, respectively.
\[(6+3i)(6-3i) = \boxed{45}\]
\[(4-5i)(4+5i) = \boxed{41}\]
\[(-3+8i)(-3-8i) = \boxed{73}\]

Examples

Complex numbers are essential in fields like electrical engineering and physics to analyze alternating current (AC) circuits, signal processing, and wave phenomena. For example, when calculating the impedance of an electronic component, you might multiply complex quantities to find the total resistance and reactance in a circuit. This process helps engineers design more efficient power systems and communication networks, ensuring everything from your phone battery to the power grid functions correctly.