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 HalfAngle 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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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