2ObjectiveSimplify rational expressions containing complex numbers in the denominator.
3Assignment Pgs. 299-300 #5-61 odd only Chapter 6 Review. Pgs # Bring your questions to class so you are prepared next time we test.Chapter 6 notes are due in your binder on test date.
4Note:As you know, two radical expressions of the form a√b + c √d and a √b – c √d are conjugates of each other. Since imaginary numbers involve radicals, numbers of the form a + bi and a – bi are also called conjugates. Recall that the product of two radical conjugates is a rational number. Let’s investigate the product of two complex conjugates.
5Ex. 1: Find the product of the conjugates 5 + 2i and 5 – 2i. (5 + 2i)(5 – 2i) = 25 – 10i + 10i -4i2= 25 – 4(-1)== 29Reminder: i2 = -1
6Ex. 2: Find the product of the conjugates a + bi and a – bi. (a + bi)(a – bi) = a2 – abi + abi - bi2= a2 –bi2= a2 – b2(-1)= a2 +b2Reminder: i2 = -1Since a and b are real, a2b2 will also be real.
7Note:To simplify expressions, we must eliminate all radicals in the denominator. Since i is a radical, all imaginary numbers must also be eliminated from denominators.Ex. 3: SimplifyIf you wanted to multiply by –i/-i, you could have.
8Rationalizing the denominator We use conjugates of radicals to rationalize the denominators of expressions with radicals in the denominator. We can also use conjugates of complex numbers to rationalize the denominators of expressions with complex numbers in the denominator.
10Ex. 5: Find the multiplicative inverse of 5 – 7i. The multiplicative inverse of 5 – 7i is
11Ex. 5: Check your work! The inverse of 5 – 7i is The multiplicative inverse of 5 – 7i isThe inverse of 5 – 7i is
12Reminder:Review over chapter 6 followed by the exam. Bring your questions to class next time we meet. We will not be doing the review in class. It is an assignment due after you turn in your test.Don’t forget your chapter 6 notes in your binder for me to check.