imaginary numbers?

Question

a) using your knowledge of algebra and imaginary numbers, solve the following equation:
z^2 + (1/2)z + (1/4) = 0

b) if w= 8i
then what is w written in polar form?

Answer 1

a) Quadratic formula, a = 1, b = 1/2, c = 1/4

[-b ± √(b² - 4ac)]/2a

[-1/2 ± √(1/4 - (4)(1)(1/4)] /2

-1/4 ± 1/2[√(1/4 - 1)]

-1/4 ± 1/2√(-3/4)

-1/4 ± (i/4)(√3)

(1/4)(1 ± i√3)

b) length is 8 (there is no real number component), direction is purely the i-axis, or π/2 ==> (8, π/2)

Answer 2

1) Use the quadratic formula for this problem.
z = \frac{-b +/- \sqrt{b^2 - 4ac}}{2a}
z = \frac{-1/2 +/- \sqrt{(1/2)^2 - 4*1*(1/4)}}{2*1}
z = \frac{-1/2 +/- \sqrt{1/4 - 1}}{2}
z = \frac{-1/2 +/- \sqrt{-3/4}}{2}
z = -1/4 +/- (i/4)*\sqrt{3}

2) Polar form has a length of \sqrt{a^2 + b^2} where the number is a+bi. So in this case the length is \sqrt{0^2 + 8^2} which is merely 8. The angle is calculated by arctan{b/a}. So in this case the angle is arctan{8/0} which is arctan{infinite} which is pi/2 radians or 90 degrees. Therefore in polar form w is (8,pi/2) or (8,90).

Answer 3

Use the resolving formula as for real numbers

z = (-b+or - sqrt(b^2 -4ac)/2a

z = [-1/2+or-sqrt(1/4 - 1)]/2

z1 = -1/2 + sqrt(-1x3/4)= -1/2 + sqrt(-1)xsqrt(3/4)=
= -1/2 + isqrt(3/4)


The other root is similarly found.



======================

w = 8i in polar form is 8Cis90º or 8sin90º better 8sinpi/2

Answer 4

Question a)
Z = [ -1/2 ± √(1/4 - 1) ] / 2

Z = [ - 1/2 ± √ (- 3 / 4) ] / 2

Z = [ - 1/2 ±√ [ i ² 3 / 4] / 2

Z = - 1 / 4 ± i .√3 / 4

Z = - 1 / 4. ( 1 ± i.√ 3 )

Question b)
w = 8 / 90° = 8 (cos 90° + i sin 90°)

Answer 5

a) 4z^2+2z+1=0 so z=((-2+-sqrt(4-16))/8 =(-1+-i sqrt(3) )/4

b)= 8

Answer 6

a) Use the quadratic formula:

-b ± √(b² - 4ac)
---------------------
2a

-½ ± √(¼ - 4(¼))
-----------------------
2

-¼ ± √(-3/16)

-¼ ± ¼√(-3)

-¼ ± ¼i√(3)

b) i is graphed on the y-axis, so this would be graphed as (8,π/2).

Answer 7

what??