Transcript HERE - KNOWING IS NOT UNDERSTANDING

Solving Complex Equations
𝑛
of the form 𝑧 = 𝑎 + 𝑖𝑏
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!
z3 = 8 or z3 = 8 + 0i
let z = rcis()
1.
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
r =2
Solutions:
z1 = 2 cis(0)
z2 = 2 cis(120)
z3 = 2 cis(240)
3 = 0 + 360n
 = 120n
= 0, 120, 240
z3 = – 8
let z = rcis()
2.
or
z3 = – 8 + 0i
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
r =2
Solutions:
z1 = 2 cis(60)
z2 = 2 cis(180)
z3 = 2 cis(30)
3 = 180 + 360n
 = 60 +120n
= 60, 180, 300
z3 = 8i or z3 = 0 + 8i
let z = rcis()
3.
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
r =2
Solutions:
z1 = 2 cis(30)
z2 = 2 cis(150)
z3 = 2 cis(270)
3 = 90 + 360n
 = 30 + 120n
= 30, 150, 270
z3 = – 8i or z3 = 0 – 8i
let z = rcis()
4.
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
r =2
Solutions:
z1 = 2 cis(90)
z2 = 2 cis(210)
z3 = 2 cis(330)
3 = 270 + 360n
 = 90 +120n
= 90, 210, 330
z3 = 3 + 3i
let z = rcis()
5.
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
r = 18 (1/6)
3 = 45 + 360n
 = 15 + 120n
= 15, 135, 255
Solutions:
z1 = 18 (1/6)cis(15)
z2 = 18 (1/6)cis(135)
z3 = 18 (1/6)cis(255)
z3 = –3 + 3i
let z = rcis()
6.
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
r = 18 (1/6)
3 = 135 + 360n
 = 45 + 120n
= 45, 165, 285
Solutions:
z1 = 18 (1/6)cis(45)
z2 = 18 (1/6)cis(165)
z3 = 18 (1/6)cis(285)
z3 = k2 or
let z = rcis()
7.
z3 = k2 + 0i
length = +k2
angle = 00
so |k2 + 0i| = +k2 and arg(k2 + 0i) = 00
r3cis (3) = k2+ 0i
r3cis (3) = +k2 cis(0 + 360n)
r3 = k2
r = k⅔
3 = 0 + 360n
 = 120n
= 0, 120, 240
Solutions:
z1 = k ⅔cis(0)
z2 = k ⅔cis(120)
z3 = k ⅔ cis(240)
8.
z4 + k2 = 0 then z4 = –k2 + 0i
let z = rcis()
length = +k2
angle = 1800
so |–k2 + 0i| = +k2 and arg(–k2 + 0i) = 1800
r4cis (4) = +k2 cis(180 + 360n)
r4 = k2
r = k½
4 = 180 + 360n
 = 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)