only the one in circle. 1.3 Complex Numbers 111 PO (b) i (a) Because i1, write t
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only the one in circle.
1.3 Complex Numbers 111 PO (b) i (a) Because i1, write the given power as a product involving (b) Multiply i - by I in the form of i to create the least positive exponent for i. Now Try Exercises 89 and 97 Exercises CONCEPT PREVIEW Fill in the blank to correctly complete each sentence 1. By definition, and therefore, i- 2. If a and b are real numbers, then any number of the form a bi is a(n) 3. The numbers 6 +5i and 6 -5i, which differ only in the sign of their imaginary parts, are 4. The product of a complex number and its conjugate is always a(n) 5. To find the quotient of two complex numbers in standard form, multiply both the numerator and the denominator by the complex conjugate of the CONCEPT PREVIEW Decide whether each statement is true or falsc. If false, correct the right side of the equation. 6, V-25 = 5i 9. (-2+7) _ (10-6) =-12+i 10. (5 +3):16 Concept Check Identify each number as real, complex, pure imaginary, or nonreal com plex. (More than one of these descriptions will apply.) 14.-7i 15/ 5+i 12.0 13. 13i 19. V-25 20. V-36 17, 18. V24 16.-6-2i Write each number as the product of a real number and i. See Example I. 21. V-25 22. V-36 23. V-10 24. V-15 25. V-288 27.-V-18 -V-80 28. 26. V-500 Find each product or quotient. Simplify the answers. See Example 2. 29. V-13 V-13 30. V-17 30 10 31. V-3 V-8 V-70 V-7 34.Explanation / Answer
1) ( -2 + 7i ) - ( 10 - 6i )
real part is combined with real part and imaginary part with imaginary
(-2 -10 )+ ( 7i - 6i )
-12 + i
2) sqrt (-4) * sqrt (-9)
sqrt (-4) = sqrt (4) * sqrt (-1)
sqrt (-1) = i
so we can write sqrt (-4) = 2i
similarly , sqrt (-9) = sqrt (3) * sqrt (-1) = 3i
now we have (2i)(3i) = 6i^2
i^2 = -1
6(-1) = -6
hence , sqrt (-4) * (sqrt (-9) = -6
11) -4 is a real and complex number
13) 13i
it is complex , pure imaginary and non real complex
15) 5 + i
complex and non real complex
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