How do I write and evaluate terms in polar form, and how can I find terms given the root of a quadratic equation.

Question

1.

a) For z = 2 + 2(√3)i

i)     How can I write z in polar form

ii)   How will I evaluate z³ in polar form

iii) and how can I find z^→ and evaluate (z^→)³

 

2.

a) If a = -3 – 2i, is a root of the quadratic equation, z² + az + b = 0,  a, b ∈ ℝ

How can I find a and b ?

Answer 1

1ai)

z = 2 + 2isqr3

graph i on the vertical, y axis, reals on the horizontal, x axis

it helps to draw at least a rough sketch

x=2

y=2sqr3

find the hypotenuse: r^2 =x^2 + y^2= 2^2 + (2sqr3)^2 = 4+12 = 16, r=sqr16 = 4

continued below after 2a)

2a)

x=-3-2i

add 3 to both sides

x+3 = -2i

square both sides

x^2+6x+9 =-4

add 4 to both sides

x^2+6x+13 = 0

for x^2+ax +b

a=6

b=13

check the answer

x=-b/2a +/-(1/2a)sqr(b^2-4ac)

but a and b are switched as well as b and c, in the version you have,so b=6, a=1, c=13

=-6/2+/-(1/2)sqr(36-4(13))

= -3+/-(i/2)sqr16

=-3+/-2i

= -3-2i, -3+2i

imaginary solutions come in conjugate pairs

1ai continued)

y=rsinT= 2sqr3

rcosT = x = 2

4cosT= 2

cosT= 2/4=1/2= .5

T = cos^-1(.5) = 60 degrees = pi/3 radians

polar coordinates are: (r, Theta) = (4, angle) = (4, pi/3)= (4,60)

z^3= r^3(cos3(60)+isin3(60)) = 64(cos180+0) = -64

Answer 2

z=2+i(2√3)=4(1/2+i√3/2)=4(cos(π/3)+isin(π/3))

z³=[4(cos(π/3)+isin(π/3)]³

use de moivre thoerem

z³=64(cos(π)+isin(π))=-64

?>what does notation mean z^-> and (z^->)³?

this one

z₁₌-3 – 2i, is a root of the quadratic equation, z² + az + b = 0,  a, b ∈ ℝ and z₁ is a value solving equation

then z₂=-3+2i if a and b real

have you been taught or see on your own

if f(z)=(z-z₁)(z-z₂)=z²+bz+c,,,,,,b=-(z₁+z₂),,,,c=z₁z₂

here b=6 and c=13