Find the product of each pair of complex conjugates.
$\begin{array}{l}
(3+8 i)(3-8 i)= \\
(4+5 i)(4-5 i)=
\end{array}$
We are asked to compute the products of two pairs of complex conjugates: \((3+8i)(3-8i)\) and \((4+5i)(4-5i)\). A complex number is written in the form \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The conjugate of this number is \(\bar{z} = a - bi\). The product of a complex number and its conjugate is a fundamental property in algebra: \[(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2.\] Since \(i^2 = -1\), the product simplifies to \(a^2 + b^2\). This identity allows us to compute these products directly.
For the first product \((3+8i)(3-8i)\), we identify the real part \(a = 3\) and the imaginary part \(b = 8\). Applying the identity \(a^2 + b^2\), we have:
\[(3+8i)(3-8i) = 3^2 + 8^2 = 9 + 64 = 73.\]
For the second product \((4+5i)(4-5i)\), we identify the real part \(a = 4\) and the imaginary part \(b = 5\). Applying the identity \(a^2 + b^2\), we have:
\[(4+5i)(4-5i) = 4^2 + 5^2 = 16 + 25 = 41.\]
We have computed the two products. The product of \((3+8i)(3-8i)\) is \(73\), and the product of \((4+5i)(4-5i)\) is \(41\).
The final results are:
\[(3+8i)(3-8i) = \boxed{73}\]
\[(4+5i)(4-5i) = \boxed{41}\]
In electrical engineering, calculating the impedance of circuits often involves complex numbers. Multiplying complex impedance by its conjugate is used to find the magnitude squared of the impedance, which relates directly to power dissipation. For example, if a component has an impedance of \(3+8i\) ohms, calculating \((3+8i)(3-8i)\) helps determine the load characteristics in a circuit analysis.