Let $b$ be a real number such that the following product is also equal to a real number:
$(-7-b i)(-7+2 i)$
What is the value of $b$?
We are given the expression \((-7 - bi)(-7 + 2i)\), where \(b\) is a real number. For a complex number \(z = x + yi\) (where \(x\) and \(y\) are real numbers), the number is real if and only if its imaginary part \(y\) is equal to \(0\). Our goal is to expand the product and determine the value of \(b\) that makes the imaginary part zero.
We expand the product \((-7 - bi)(-7 + 2i)\) using the distributive property (often called the FOIL method):
\[(-7 - bi)(-7 + 2i) = (-7)(-7) + (-7)(2i) + (-bi)(-7) + (-bi)(2i)\]
Calculating each term individually:
Since \(i^2 = -1\), the last term becomes \(-2b(-1) = 2b\). Combining these, we have:
\[49 - 14i + 7bi + 2b\]
Grouping the real and imaginary parts gives:
\[(49 + 2b) + (-14 + 7b)i\]
For the product to be a real number, the coefficient of the imaginary part must be zero. The imaginary part is \((-14 + 7b)\). Setting this equal to zero, we get the equation:
\[-14 + 7b = 0\]
Adding \(14\) to both sides, we get:
\[7b = 14\]
Dividing by \(7\), we find:
\[b = 2\]
Thus, when \(b = 2\), the product becomes \((49 + 2(2)) + (-14 + 7(2))i = 53 + 0i = 53\), which is a real number.
In electronics, complex numbers are used to represent impedance in AC circuits. The impedance is a complex quantity where the real part represents resistance and the imaginary part represents reactance. To ensure that a circuit component behaves purely as a resistor (meaning its total impedance is a real number), engineers must adjust circuit parameters—like the values of inductors or capacitors—so that the total reactance (the imaginary part) becomes zero. This problem demonstrates the fundamental algebraic process of balancing these components.