PURE MATHEMATICS

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1. The complex number 'a' is such that a^2=5-12i
use an algebraic method to find the two possible values of 'a'

8. The complex number z is given by z=cosTheta + isinTheta where -1/2pi< or equal to Theta less than or equal to 1/2pi.

a. show that |1+z|=2cos(1/2Theta) and arg(1+z)=1/2Theta.

b. hence, show that 1/(1+z)=1/2[1-itan(1/2Theta)]

c. describe the locus of the point representing z and the locus of the point representing 1/(1+z) in an argand diagram as Theta varies from -pi/2 to pi/2

9. A) find the exact value of all the roots of the equation (z+2+5i)(z^2+2z+5)=0

  • PURE MATHEMATICS -

    1. Let a = x + yi for reals x and y

    Thus solve the simultaneous equations:
    x^2-y^2 = 5, and 2xy = -12

    8. 1+z = 1+cos(θ) + i sin(θ)
    Hint: Substitute θ=2ω then simplify
    Use: sin(2ω) = 2 sin(ω) cos(ω)
    And: cos(2ω) = 2 cos^2(ω) - 1

    9. One root: z = -2-5i
    The other roots will be those of:
    z^2 + 2z + 5 = 0
    Hint: Use either complete the square, or the quadratic formula.

  • PURE MATHEMATICS -

    #1
    since 5 = 3^2-2^2,
    (3-2i)^2 = 9-12i-4 = 5-12i
    (-3+2i)^2 = 5-12i

    Or, you can do it by equating real/imaginary parts:

    (a+bi)^2 = (a^2-b^2) + 2abi
    so
    a^2-b^2 = 5
    2ab = 12
    a^2 - (6/a)^2 = 5
    and solve for a

    #8 I don't think this is true.
    If z = 1+i, θ=pi/4
    1+z = 2+i
    |1+z|=√5 and arg(z+1) = arctan(1/2) which is not pi/8

    #9
    z^2+2z+5 = (z+1)^2 + 2^2 = (z+1+2i)(z+1-2i)
    so,
    z = -2-5i or -1-2i or -1+2i

  • PURE MATHEMATICS -

    for the first one I got a to be -0.67 and 1.5 when b is 9 and -4, respectively

  • PURE MATHEMATICS -

    Steve, #8 is true. Your example is wrong.
    Note: if θ=(π/4) then z = (1/√2) + (i/√2)

  • PURE MATHEMATICS -

    Hmmm. Makes sense. Thanks for checking. I forgot we were on the unit circle.

  • PURE MATHEMATICS -

    so number 8 is has trigonometry identities in it?

  • PURE MATHEMATICS -

    thanks very much for helping with 1 and 9
    number 8 is still giving me some trouble though

  • PURE MATHEMATICS -

    #8. 1+z = 1+cos(θ) + i sin(θ)
    Substitute θ=2ω then simplify
    Use: sin(2ω) = 2 sin(ω) cos(ω)
    And: cos(2ω) = 2 cos^2(ω) - 1

    So:
    1+z = 1+cos(2ω) + i sin(2ω)
    1+z = 2 cos^2(ω) + 2i sin(ω) cos(ω)
    1+z = 2 cos(ω) (cos(ω) + i sin(ω))
    1+z = 2 cos(θ/2)(cos(θ/2) + i sin(θ/2))

    Thus for -π/2 ≤ θ ≤ π/2:
    |1+z| = 2 cos(θ/2)
    arg(1+z) = θ/2


    Then for (b) use:
    1/(cos(ω)+i sin(ω)) = cos(ω)-i sin(ω)
    To find 1/(1+z)


    Then for (c) plot z and 1/(1+z) on the complex plane and describe.

  • PURE MATHEMATICS -

    Explain to me just this one line 1+z = 2 cos(θ/2)(cos(θ/2) + i sin(θ/2))what happened to i sin(θ/2)) in the other two lines which follow?

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