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    week 2 worksheet answers

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    week 2 worksheet answers

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    week 2 worksheet answers

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    ayushdash78

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    MATH 211 Fall 2024

    Week 2 (9 Sep) worksheet Solutions


    You must clearly show your steps for every problem below.

    1. Factorize z3 + 27i in linear polynomials.


    Sol. The key observation here is that (x − α) is a linear factor of a polynomial f(x) if and
    only if α is a solution of the polynomial equation f(x) = 0, i.e. f(α) = 0. For example,
    x2 − 5x + 6 = 0 has solutions x = 2, 3 and so x2 − 5x + 6 = (x − 2)(x − 3). So to
    factorize in linear factors, we want to find all the solutions of
    z3 + 27i = 0
    ⇒ z3 = 27(−i)
    i(− π +2kπ)
    ⇒ z3 = 27e 2 where k ∈ Z

    i(− π + 2kπ )
    ⇒ z = 3e 6 3

    − πi πi 7πi
    Choosing k = 0,√1, 2 gives us the √
    three distinct roots 3e 6 , 3e 2 , 3e 6 , in

    Cartesian form 3 2 3 − 3i2 , 3i, − 3 2 3 − 3i2 So the factorization is


    √ ! √ !
    3 3 3i 3 3 3i
    z3 + 27i = (z − 3i) z − + z+ +
    2 2 2 2
    √ √
    Other version answers : z√3 − 8i = (z + √2i)(z − 3 − i)(z + 3 − i) and
    3
    z + 8i = (z − 2i)(z + i − 3)(z + i + 3).
    2. Find all solutions of the quadratic equation
    z2 + (6 − 4i)z + 10 = 0.

    Sol. Using the quadratic formula,


    p
    −(6 − 4i) ± (6 − 4i)2 − 40
    z=
    2
    p √
    Now, (6 − 4i)2 − 40 = 2 −5 − 12i. To find this square root, suppose
    √
    −5 − 12i = a + bi ⇒ −5 − 12i = a2 − b2 + 2abi.
    Comparing real and imaginary parts, we need to solve a2 − b2 = −5 and ab = −6.
    Substituting b = − a6 in the first we get a2 − a362 = −5 ⇒ a4 + 5a2 − 36 = 0. Using the
    quadratic formula again, a2 = −5±13 2
    . Note that a is a real number, so a2 ≥ 0. Thus,
    a = 4 ⇒ a = ±2. In fact, from ab = −6, we see that when a = 2, b = −3 and when
    2

    a = −2, b = 3. Thus,
    −(6 − 4i) ± 2(2 − 3i)
    z= ⇒ z = −1 − i, −5 + 5i
    2
    Other version answers : For z2 + (−6 + 2i)z + (11 − 2i) = 0, z = 2 + i, 4 − 3i.
    For z2 − (3 + 3i)z + 29i = 0, z = −2 + 5i, 5 − 2i

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