Stitz-Zeager_College_Algebra_e-book

141 parabola axis of symmetry 141 denition 407 focal

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Unformatted text preview: 2 3 √ π (b) z = 7cis − 34 = − 7 2 2 − √ 3. Since z = − 3 2 3 + 3 i = 3cis 2 √ 72 2i 5π 6 7π 12 − 11π 12 (a) zw = 18cis (b) z w 1 = 2 cis (d) π 2 w0 = cis π 6 √ = 3 2 4 3 = 3 + 4i √ √ and w = 3 2 − 3i 2 = 6cis − π , we have the following. 4 (c) 4. (a) Since z = i = cis (d) z = 5cis arctan 11π w z = 2cis 12 π z 4 = 81cis − 23 π (e) w3 = 216cis − 34 (f) z 5 w2 = 8748cis − π 3 we have 1 + 2i w1 = cis 5π 6 √ 3 2 =− + 1i 2 w2 = cis 3π 2 = −i (b) Since z = 64 = 64cis(0) we have w0 = 2cis(0) = 2 √ 2π 3 π w3 = 2cis (π ) 3 = 1 + 3i √ Since 5 − 5 2 3 i = 5cis − π we have 2 3 √ √ √ 15 w0 = 5cis − π = 2 − 25 i 6 w1 = 2cis (c) w2 = 2cis = −1 + √ √ π w4 = 2cis − 23 = −1 − 3i √ w5 = 2cis − π = 1 − 3i 3 3i = −2 w1 = √ 5cis 5π 6 √ =− 15 2 √ + 5 2i 5. Note: In the answers for w0 and w2 the first rectangular form comes from applying the π π π appropriate Sum or Difference Identity ( 12 = π − π and 17π = 23 + 34 , respectively) and the 3 4 12 second comes from using the Half-Angle Identities. √√ √√ √√ √ √ √ √6+√2 2+ 3 3 3 3 6− 2 π w0 = 2cis 12 = 2 +i =2 + i 2− 3 4 4 2 2 w1 = w2 = √ 3 √ 3 2cis 3π 4 2cis 17π 12 = √ 3 = √ √ 2− √ 3 √ 2 2 2 + √ 2− 6 4 2 2i +i √√ − 2− 6 4 = √ 3 √ 2 √ 2− 3 2 √ +i √ 2+ 3 2 858 Applications of Trigonometry 6. w0 = cis(0) = 1 w1 = cis w2 = cis w3 = cis w4 = cis 2π 5 4π 5 6π 5...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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