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    i seem to be asking a lot of questions,
    but its only questions that i have
    and answers short of.

    i have appended my p6 problems. all of ur help wud be appreciated :adore: . thank you
    have a great day.
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  1. File Type: pdf P6 complex nos..pdf (52.1 KB, 147 views)
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    70

    Let z = x+iy. Then

    Then Im[(2z+1)/(iz+1)]

    = Im[(2x+1+2iy)/(1-y+ix)]

    = Im[(2x+1+2iy)(1-y-ix)/((1-y)^2+x^2)] <multiply by conjugate of the denominator>

    = [2y(1-y) - x(2x+1)]/[1-2y+y^2+x^2]

    So Im[...] = -2 if

    2y - 2y^2 - 2x^2 - x = -2 +4y - 2y^2 - 2x^2

    2 = x +2y

    Which is the equation of a line
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    72

    As |z-3| = 2 is a circle, centre 3, radius 2, then |z|=5 is the maximum, attained at z = 5
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    Q48(Page2)
    w=z^{2}
    z describes y=1 so z=a+i.
    w=(a+i)^{2}\\

w=a^{2}+2ai-1\\

w=[a^{2}-1]+i[2a]
    If w=u+iv\\

u=[a^{2}-1], v=2a\\

a^{2}=u+1=\frac{v^{2}}{4}\\

4(u+1)=v^{2}\\

v^{2}=4(u+1)
    This is, of course, a parabola of the form v^{2}=4au but translated one unit in the direction of the negative x axis.
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    112

    (a) The three cube roots of -64 are

    -4

    4 (cos(pi/3) + i sin(pi/3))

    4 (cos(2pi/3) + i sin(2pi/3)

    and they sit equally spaced around the circle |z|=4

    (b) The sixth roots of -64 are

    2 [cos ((2n+1) pi/6) + i sin((2n+1)pi/6)]

    for n = 0, 1, 2, 3, 4, 5

    and they sit equally spaced around the circle |z|=2
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    34

    The map W = 2(Z+ 3 - i)

    maps the unit circle to the circle with centre 3-i and radius 2

    If w = (1-z)/z maps the line y = 1/2 to the unit circle

    Then the combination of the two will map the line y = 1/2 to this second circle - and this equals

    (- 2Z - 5 + 2i)/(2Z + 6 - 2i)
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    sin7t = 7 sint

    if (from the formular for sint7t)

    -64 (sint)^7 + 112 (sint)^5 - 56 (sint)^3 = 0 (*)

    If sint =/= 0 then

    -64 (sint)^4 + 112 (sint)^2 - 56 =0

    Write z = (sint)^2

    -64 z^2 + 112 z - 56.

    8z^2 - 14z + 7 = 0

    Now this has discriminatnt

    14^2 - 4.7.8 = -28

    and so it has no real roots.

    So the equation (*) has real roots for sint only when sint=0 i.e. when t = n pi
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    w = (1+3z)/(1-z)

    Find the image of Rez = 0

    Method 1: a general point on the line is iy

    (1+3iy)/(1-iy) = (1+3iy)(1+iy)/(1+y^2)

    So

    u = (1 -3y^2)/(1+y^2) = 1 - 4y^2/(1+y^2)
    v = (4y)/(1+y^2)

    Then (1-u)/v = y

    Substituting into the v-equation:

    v [1 + (1-u)^2/v^2] = 4(1-u)/v

    v^2 + 1 - 2u + u^2 = 4 - 4u

    u^2 +2u +v^2 = 3

    (u+1)^2 + v^2 = 4

    Method 2:

    (Let z* denote z conjugate in the following - note for complex numbers (ab)* = a*b* and (a/b)* = a*/b* and |a|^2 = aa*)

    w = (1+3z)/(1-z)

    w - wz = 1 + 3z

    z(3+w) = w - 1

    z = (w-1)/(3+w)

    Now w is on the image C only when z is on the imaginary axis - this has equation z = - z*

    So w is on C precisely when

    (w-1)/(3+w) = - [(w-1)/(3+w)]* = - (w*-1)/(3+w*)

    So

    (w-1)(3+w*) = - (w*-1)(3+w)

    -3 + 3w - w* + ww* = 3 - 3w* + w - ww*

    ww* +w* + w - 3 = 0

    (w+1)(w*+1) = 4

    |w+1|^2 = 4

    |w+1| = 2

    as required
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    (Original post by RichE)
    70

    Let z = x+iy. Then

    Then Im[(2z+1)/(iz+1)]

    = Im[(2x+1+2iy)/(1-y+ix)]

    = Im[(2x+1+2iy)(1-y-ix)/((1-y)^2+x^2)] <multiply by conjugate of the denominator>

    = [2y(1-y) - x(2x+1)]/[1-2y+y^2+x^2]

    So Im[...] = -2 if

    2y - 2y^2 - 2x^2 - x = -2 +4y - 2y^2 - 2x^2

    2 = x +2y

    Which is the equation of a line
    oh thank you thank you thank you thank you...for all your help. u saved the day. hurray :party:
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    (Original post by Gaz031)
    Q48(Page2)
    w=z^{2}
    z describes y=1 so z=a+i.
    w=(a+i)^{2}\\

w=a^{2}+2ai-1\\

w=[a^{2}-1]+i[2a]
    If w=u+iv\\

u=[a^{2}-1], v=2a\\

a^{2}=u+1=\frac{v^{2}}{4}\\

4(u+1)=v^{2}\\

v^{2}=4(u+1)
    This is, of course, a parabola of the form v^{2}=4au but translated one unit in the direction of the negative x axis.
    thank you thank you.....
 
 
 
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