2. 1. z 3 = 8 or z 3 = 8 + 0i let z = rcis(  ) length = 8 angle = 0 0 so |8+0i| = 8 and arg(8+0i) = 0 0 r 3 cis (3  ) = 8 + 0i r 3 cis (3  ) = 8 cis(0 + 360n) r 3 = 8 3  = 0 + 360n r = 2  = 120n = 0, 120, 240 Solutions: z 1 = 2 cis(0) z 2 = 2 cis(120) z 3 = 2 cis(240)

3. 2. z 3 = – 8 or z 3 = – 8 + 0i let z = rcis(  ) length = +8 angle = 180 0 so | – 8 + 0i | = +8 and arg(– 8 + 0i ) = 180 0 r 3 cis (3  ) = 8 + 0i r 3 cis (3  ) = +8 cis(180 + 360n) r 3 = 8 3  = 180 + 360n r = 2  = 60 +120n = 60, 180, 300 Solutions: z 1 = 2 cis(60) z 2 = 2 cis(180) z 3 = 2 cis(30)

4. 3. z 3 = 8i or z 3 = 0 + 8i let z = rcis(  ) length = +8 angle = 90 0 so |0 + 8i| = +8 and arg(0 + 8i) = 90 0 r 3 cis (3  ) = 0 + 8i r = +8  = 90 0 r 3 cis (3  ) = +8 cis(90 + 360n) r 3 = 8 3  = 90 + 360n r = 2  = 30 + 120n = 30, 150, 270 Solutions: z 1 = 2 cis(30) z 2 = 2 cis(150) z 3 = 2 cis(270)

5. 4. z 3 = – 8i or z 3 = 0 – 8i let z = rcis(  ) length = +8 angle = 270 0 so| 0 – 8i| = +8 and arg(0 – 8i ) = 270 0 r 3 cis (3  ) = 0 – 8i r = +8  = 270 0 r 3 cis (3  ) = +8 cis(270 + 360n) r 3 = 8 3  = 270 + 360n r = 2  = 90 +120n = 90, 210, 330 Solutions: z 1 = 2 cis(90) z 2 = 2 cis(210) z 3 = 2 cis(330)

6. 5. z 3 = 3 + 3i let z = rcis(  ) length = √(9 + 9)= √18 angle = 45 0 so |3+3i| = 18 ½ and arg(3+3i) = 45 0 r 3 cis (3  ) = (18 ½ )cis( 45 + 360n) r 3 = 18 ½ 3  = 45 + 360n r = 18 (1/6)  = 15 + 120n = 15, 135, 255 Solutions: z 1 = 18 (1/6) cis(15) z 2 = 18 (1/6) cis(135) z 3 = 18 (1/6) cis(255)

7. 6. z 3 = –3 + 3i let z = rcis(  ) length = √(9 + 9)= √18 angle = 135 0 so | –3 + 3i | = 18 ½ and arg(–3 + 3i ) = 135 0 r 3 cis (3  ) = (18 ½ )cis( 135 + 360n) r 3 = 18 ½ 3  = 135 + 360n r = 18 (1/6)  = 45 + 120n = 45, 165, 285 Solutions: z 1 = 18 (1/6) cis(45) z 2 = 18 (1/6) cis(165) z 3 = 18 (1/6) cis(285)

8. 7. z 3 = k 2 or z 3 = k 2 + 0i let z = rcis(  ) length = +k 2 angle = 0 0 so |k 2 + 0i| = +k 2 and arg(k 2 + 0i) = 0 0 r 3 cis (3  ) = k 2 + 0i r 3 cis (3  ) = +k 2 cis(0 + 360n) r 3 = k 2 3  = 0 + 360n r = k ⅔  = 120n = 0, 120, 240 Solutions: z 1 = k ⅔ cis(0) z 2 = k ⅔ cis(120) z 3 = k ⅔ cis(240)

9. 8. z 4 + k 2 = 0 then z 4 = –k 2 + 0i let z = rcis(  ) length = +k 2 angle = 180 0 so |–k 2 + 0i| = +k 2 and arg(–k 2 + 0i) = 180 0 r 4 cis (4  ) = +k 2 cis(180 + 360n) r 4 = k 2 4  = 180 + 360n r = k ½  = 45 + 90n = 45, 135, 225, 315 Solutions: z 1 = k ½ cis(45) z 2 = k ½ cis(135) z 3 = k ½ cis(225) z 4 = k ½ cis(315)