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Solving Complex Equations of the form IMPORTANT POINT:

Solving Complex Equations of the form IMPORTANT POINT: If z = r cis( ) then the value of r can only be a positive number. It is the length of the complex number!. 1. z 3 = 8 or z 3 = 8 + 0i let z = rcis (  )

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Solving Complex Equations of the form IMPORTANT POINT:

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  1. Solving Complex Equations of the form IMPORTANT POINT: If z = r cis() then the value of r can only be a positive number. It is the length of the complex number!

  2. 1. z3 = 8 or z3 = 8 + 0i let z = rcis() length = 8 angle = 00 so |8+0i| = 8 and arg(8+0i)= 00 r3cis (3) = 8 + 0i r3cis (3) = 8 cis(0 + 360n) r3 = 8 3 = 0 + 360n r = 2  = 120n = 0, 120, 240 Solutions: z1 = 2 cis(0) z2 = 2 cis(120) z3 = 2 cis(240)

  3. 2. z3 = – 8 or z3 = – 8 + 0i let z = rcis() length = +8 angle = 1800 so | – 8 + 0i | = +8 and arg(– 8 + 0i ) = 1800 r3cis (3) = 8 + 0i r3cis (3) = +8 cis(180 + 360n) r3 = 8 3 = 180 + 360n r = 2  = 60 +120n = 60, 180, 300 Solutions: z1 = 2 cis(60) z2 = 2 cis(180) z3 = 2 cis(30)

  4. 3. z3= 8i or z3 = 0 + 8i let z = rcis() length = +8 angle = 900 so |0 + 8i| = +8 and arg(0 + 8i) = 900 r3cis (3) = 0 + 8i r = +8  = 900 r3cis (3) = +8 cis(90 + 360n) r3 = 8 3 = 90 + 360n r = 2  = 30 + 120n = 30, 150, 270 Solutions: z1 = 2 cis(30) z2 = 2 cis(150) z3 = 2 cis(270)

  5. 4. z3 = – 8i or z3 = 0 – 8i let z = rcis() length = +8 angle = 2700 so| 0 – 8i| = +8 and arg(0 – 8i ) = 2700 r3cis (3) = 0 – 8i r = +8  = 2700 r3cis (3) = +8 cis(270 + 360n) r3 = 8 3 = 270 + 360n r = 2  = 90 +120n = 90, 210, 330 Solutions: z1 = 2 cis(90) z2 = 2 cis(210) z3 = 2 cis(330)

  6. 5. z3= 3 + 3i let z = rcis() length = √(9 + 9)= √18 angle = 450 so |3+3i| = 18 ½ and arg(3+3i)= 450 r3cis (3) = (18 ½)cis( 45 + 360n) r3 = 18 ½ 3 = 45 + 360n r = 18 (1/6) = 15 + 120n = 15, 135, 255 Solutions: z1 = 18 (1/6)cis(15) z2 = 18 (1/6)cis(135) z3 = 18 (1/6)cis(255)

  7. 6. z3 = –3 + 3i let z = rcis() length = √(9 + 9)= √18 angle = 1350 so |–3 + 3i | = 18 ½ and arg(–3 + 3i )= 1350 r3cis (3) = (18 ½)cis( 135 + 360n) r3 = 18 ½ 3 = 135 + 360n r = 18 (1/6) = 45 + 120n = 45, 165, 285 Solutions: z1 = 18 (1/6)cis(45) z2 = 18 (1/6)cis(165) z3 = 18 (1/6)cis(285)

  8. 7. z3= k2 orz3 = k2 + 0i let z = rcis() length = +k2 angle = 00 so |k2 + 0i| = +k2and arg(k2 + 0i)= 00 r3cis (3) = k2+ 0i r3cis (3) = +k2 cis(0 + 360n) r3 = k2 3 = 0 + 360n r = k⅔ = 120n = 0, 120, 240 Solutions: z1 = k⅔cis(0) z2 = k⅔cis(120) z3 = k⅔ cis(240)

  9. 8. z4 + k2 = 0 then z4 = –k2 + 0i let z = rcis() length = +k2 angle = 1800 so |–k2 + 0i| = +k2andarg(–k2 + 0i)= 1800 r4cis (4) = +k2 cis(180 + 360n) r4 = k2 4 = 180 + 360n r = k ½  = 45 + 90n = 45, 135, 225, 315 Solutions: z1 = k ½ cis(45) z2 = k ½ cis(135) z3 = k ½cis(225) z4 = k ½cis(315)

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