UNIT 4 EXTENDING THE NUMBER SYSTEM Lesson 3: Operating with Complex Numbers Instruction

Transcription

1 Prerequisite Skills This lesson requires the use of the following skills: finding the product of two binomials simplifying powers of i adding two fractions with different denominators (for application problems) Introduction The product of two complex numbers is found using the same method for multiplying two binomials. As when multiplying binomials, both terms in the first complex number need to be multiplied by both terms in the second complex number. The product of the two binomials x + y and x y is the difference of squares: x 2 y 2. If y is an imaginary number, this difference of squares will be a real number since i i = : (a + bi )(a bi ) = a 2 (bi ) 2 = a 2 b 2 ( ) = a 2 + b 2. Key Concepts Simplify any powers of i before evaluating products of complex numbers. In the following equations, let a, b, c, and d be real numbers. Find the product of the first terms, outside terms, inside terms, and last terms. Note: The imaginary unit i follows the product of real numbers. (a + bi ) (c + di ) = ac (product of the first terms) + adi (product of the outside terms) + bci (product of the inside terms) + bidi (product of the last terms) = ac + adi + bci + bdi 2 = ac + bd( ) + adi + bci = (ac bd ) + (ad + bc)i ac bd is the real part of the product, and ad + bc is the multiple of the imaginary unit i in the imaginary part of the product. A complex conjugate is a complex number that when multiplied by another complex number produces a value that is wholly real. The product of a complex number and its conjugate is a real number. The complex conjugate of a + bi is a bi, and the complex conjugate of a bi is a + bi. U4-94

2 The product of a complex number and its conjugate is the difference of squares, a 2 (bi) 2, which can be simplified. a 2 b 2 i 2 = a 2 b 2 ( ) = a 2 + b 2 Common Errors/Misconceptions incorrectly finding the product of two complex numbers incorrectly identifying the complex conjugate of a + bi as a value such as a + bi, a + bi, or a bi U4-95

3 Guided Practice Example Find the result of i 2 5i.. Simplify any powers of i. 2 3 i i= i = i 2. Multiply the two terms. Simplify the expression, if possible, by simplifying any remaining powers of i or combining like terms. 5( i) = 5i Example 2 Find the result of (7 + 2i )(4 + 3i ).. Multiply both terms in the first polynomial by both terms in the second polynomial. Find the product of the first terms, outside terms, inside terms, and last terms. (7 + 2i )(4 + 3i ) = i + 2i 4 + 2i 3i 2. Evaluate or simplify each expression i + 2i 4 + 2i 3i = i + 8i + 6i 2 = i + 8i + 6( ) = i + 8i 6 3. Combine any real parts and any imaginary parts i + 8i 6 = i + 8i = i U4-96

4 Example 3 Find the complex conjugate of 5 i. Use multiplication to verify your answer.. Find the complex conjugate. The complex conjugate of a number of the form a bi is a + bi; therefore, the complex conjugate of 5 i is 5 + i. 2. Multiply both terms in the first polynomial by both terms in the second polynomial. Find the product of the first terms, outside terms, inside terms, and last terms. (5 i )(5 + i ) = i 5i i 2 = 25 i 2 = 25 ( ) = Verify that the product of the complex number and its conjugate contains no imaginary units, i. The product of (5 i ) and (5 + i ) is 26, which is a real number. (5 + i ) is the complex conjugate of (5 i ). U4-97

5 Example 4 A parallel circuit has multiple pathways through which current can flow. The following diagram of a circuit contains two elements, and 2, in parallel. 2 The impedance of an element can be represented using the complex number V + Ii, where V is the element s voltage and I is the element s current in milliamperes. If two elements are in a circuit in parallel, the total impedance is the sum of the reciprocals of each impedance. If the impedance of element is Z, and the impedance of element 2 is Z 2, the total impedance of the two elements in parallel is +. Z Z 2 Element has a voltage of 0 volts and a current of 3 milliamperes. Element 2 has a voltage of 5 volts and a current of 2 milliamperes. What is the total impedance of the circuit? Leave your result as a fraction.. Write each impedance as a complex number. Impedance = V + Ii Element : 0 + 3i Element 2: 5 + 2i 2. Find the reciprocal of each impedance. Element : 0+ 3i Element 2: 5+ 2i U4-98

6 3. Find a common denominator of the two fractions. Find the product of the two denominators. This will be the common denominator. (0 + 3i )(5 + 2i ) = i + 45i + 6i 2 = = Find an equivalent fraction for each element with a common denominator. Multiply the numerator and denominator of the fraction for element by the denominator of element 2. Use the product from step 3 to simplify (0 + 3i )(5 + 2i ) i + i = i 5+ 2i Multiply the numerator and denominator of the fraction for element 2 by the denominator of element. Use the product from step 3 to simplify (0 + 3i )(5 + 2i ) i 0+ 3 = + i i i 5. Find the total impedance by summing the reciprocals of the impedance of each element. 0+ 3i i 5+ 2i 0+ 3i = i+ 3i = 25+ 5i = U4-99