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Lv 4
? asked in Science & MathematicsMathematics · 5 years ago

How can I solve the following complex equations?

z^2=5+12i

z^3=1+i

z^2+(1+4i)z-3+3i=0

3 Answers

Relevance
  • 5 years ago
    Favorite Answer

    z² = 5 + 12i → imagine that: z = a + ib

    (a + ib²) = 5 + 12i

    a² + 2abi + i²b² = 5 + 12i → recall: i² = - 1

    a² + 2abi - b² = 5 + 12i

    (a² - b²) + 2abi = 5 + 12i → you compare both sides

    a² - b² = 5 → a² = 5 + b² ← equation (1)

    2ab = 12 → ab = 6 → a²b² = 36 ← equation (2)

    You restart from (2)

    a²b² = 36 → recall (1): a² = 5 + b²

    (5 + b²).b² = 36

    5b² + b⁴ = 36

    b⁴ + 5b² = 36 → let: B = b² → where B ≥ 0

    B² + 5B = 36

    B² + 5B + (5/2)² = 36 + (5/2)²

    B² + 5B + (5/2)² = (144/4) + (25/4)

    B² + 5B + (5/2)² = 169/4

    [B + (5/2)]² = (± 13/2)²

    B + (5/2) = ± (13/2)

    B = - (5/2) ± (13/2)

    B = (- 5 ± 13)/2

    B₁ = (- 5 + 13)/2 → B₁ = 8/2 → B₁ = 4 ← possible because the condition

    B₂ = (- 5 - 13)/2 → B₂ = - 18/2 → B₂ = - 9 ← no possible because the condition

    You keep only B₁ = 4 → recall: b² = B

    b² = 4

    b² = (± 2)²

    b = ± 2

    First case: b = 2 → recall: ab = 6 → a = 6/b → a = 3 → recall: z = a + ib

    → z₁ = 3 + 2i

    Second case: b = - 2 → recall: ab = 6 → a = 6/b → a = - 3 → recall: z = a + ib

    → z₂ = - 3 - 2i

    z³ = 1 + i

    m = √[(1)² + (1)²] = √2 ← this is the modulus

    tan(α) = 1/1 = 1 → α = π/4 ← this is the argument

    The modulus of z is: ³√m → i.e. = ³√(√3) = (√2)^(1/3) = [2^(1/2)]^(1/3) = 2^(1/6)

    The argument of z is: α/3 → (π/4)/3 = π/12

    The first root is:

    z₁ = 2^(1/6) * [cos(π/12) + i.sin(π/12)] = 2^(1/6) * e^[i.(π/12)] ← Euler's form

    …and to get the other roots, you add: (2π/3)

    z₂ = 2^(1/6) * e^[i.(π/12) + i.(2π/3)] = 2^(1/6) * e^[i.(3π/4)]

    z₃ = 2^(1/6) * e^[i.(3π/4) + i.(2π/3)] = 2^(1/6) * e^[i.(17π/12)]

    So, you can say that:

    z₁ = 2^(1/6) * [cos(π/12) + i.sin(π/12)]

    z₂ = 2^(1/6) * [cos(3π/4) + i.sin(3π/4)]

    z₃ = 2^(1/6) * [cos(17π/12) + i.sin(17π/12)]

    z² + (1 + 4i).z - 3 + 3i = 0

    z² + (1 + 4i).z + (- 3 + 3i) = 0

    Polynomial like: ax² + bx + c, where:

    a = 1

    b = (1 + 4i)

    c = (- 3 + 3i)

    Δ = b² - 4ac (discriminant)

    Δ = (1 + 4i)² - 4.(- 3 + 3i)

    Δ = 1 + 8i + 16i² + 12 - 12i → recall: i² = - 1

    Δ = 1 + 8i - 16 + 12 - 12i

    Δ = - 3 - 4i ← memorize this result as (i)

    Now let's calculate a complex number δ = a + ib such as: δ² = - 3 - 4i

    (a + ib)² = - 3 - 4i

    a² + 2abi + i²b² = - 3 - 4i

    a² - b² + 2abi = - 3 - 4i → you compare both sides

    a² - b² = - 3 → a² = b² - 3 ← equation (1)

    2ab = - 4 → ab = - 2 → a²b² = 4 ← equation (2)

    You restart from (2)

    a²b² = 4 → recall (1): a² = b² - 3

    (b² - 3).b² = 4

    b⁴ - 3b² = 4 → let: B = b² → where B ≥ 0

    B² - 3B = 4

    B² - 3B + (3/2)² = 4 + (3/2)²

    B² - 3B + (3/2)² = 25/4

    [B - (3/2)]² = (± 5/2)²

    B - (3/2) = ± (5/2)

    B = (3/2) ± (5/2)

    B = (3 ± 5)/2

    B₁ = (3 + 5)/2 → B₁ = 8/2 → B₁ = 4 ← possible because the condition

    B₂ = (3 - 5)/2 → B₂ = - 2/2 → B₂ = - 1 ← no possible because the condition

    You keep only B₁ = 4 → recall: b² = B

    b² = 4

    b² = (± 2)²

    b = ± 2

    First case: b = 2 → recall: ab = - 2 → a = - 2/b → a = - 1 → recall: δ = a + ib

    → δ₁ = - 1 + 2i

    Second case: b = - 2 → recall: ab = - 2 → a = - 2/b → a = 1 → recall: δ = a + ib

    → δ₂ = 1 - 2i = - δ₁

    You can see that: (δ₂)² = (δ₁)² = Δ = - 3 - 4i

    z₁ = (- b - √Δ)/2a = [- (1 + 4i) - (- 1 + 2i)]/2 = [- 1 - 4i + 1 - 2i]/2 = - 6i/2 = - 3i

    → z₁ = - 3i

    z₂ = (- b + √Δ)/2a = [- (1 + 4i) + (- 1 + 2i)]/2 = [- 1 - 4i - 1 + 2i]/2 = (- 2 - 2i)/2 = - 1 - i

    → z₂ = - 1 - i

    You can check it them if you want it

    = z² + (1 + 4i).z - 3 + 3i → when: z = - 3i

    = (- 3i)² + (1 + 4i).(- 3i) - 3 + 3i

    = 9i² - 3i - 12i² - 3 + 3i → where: i² = - 1

    = - 9 - 3i + 12 - 3 + 3i

    = 0

    = z² + (1 + 4i).z - 3 + 3i → when: z = - 1 - i

    = (- 1 - i)² + (1 + 4i).(- 1 - i) - 3 + 3i

    = 1 + 2i + i² - 1 - i - 4i - 4i² - 3 + 3i → where: i² = - 1

    = 1 + 2i - 1 - 1 - i - 4i + 4 - 3 + 3i

    = 0

  • ?
    Lv 7
    5 years ago

    There will be two square roots of course. You might be able to spot that

    z^2 = 5 + 12i = (3^2 – 2^2) + 2*3*2*i so that comparing with

    z^2 = a^2 – b^2 + 2abi, yields

    z = 3 + 2i or z = -3 – 2i

    Or you could just use algebra, substituting b = 6/a into

    a^2 – b^2 – 5 = 0 leading to

    (a^2 – 9)(a^2 + 4) = 0 and since a must be real

    a = +3 or – 3, b = +2 or - 2

    The polar approach is more useful with the second question.

    There will be 3 cube roots at equiangular spacing around the origin.

    z^3 = 1 + i = √(2)e^(iπ/4) = (r^3) e^(i3t)

    The first is 2^(1/6) e^(iπ/12) and since π(1/12 + 2/3) = 3π/4

    The next is 2^(1/6) e^(i3π/4) and using π(3/4 + 2/3) = 17π/12 = - 7π/12

    The third is 2^(1/6) e^(-i7π/12). These are acceptable answers.

    There is no simplification using exact rectangular forms, because, for example

    The first is 2^(1/6) multiplied by all of {[√(3) + 1] √(2) + {[√(3) - 1] √(2)i}/4

    Using a calculator we may show the approximate rectangular forms

    The first is ~ 1.08421508149 + 0.290514555507 i

    The next is ~ -0.793700525984 + 0.793700525984 i

    The third is ~ - 0.290514555507 - 1.08421508149 i

    z^2 + (1 + 4i)z - 3 + 3i = 0 Quadratic rules apply so we immediately have,

    Sum of roots u + v = -(1 + 4i) = -(1 + i) – 3i

    Product of roots uv = 3i - 3 = 3i * (1 + i) = -3i * -(1 + i)

    z^2 + (1 + 4i)z - 3 + 3i = (z + 3i)[z + (1 + i)] = 0

    z = -3i or z = -(1 + i)

  • ?
    Lv 7
    5 years ago

    5+12i = z^2 = (x+yi)^2 = (x^2-y^2) + (2xy)i, so x^2 - y^2 = 5 and 2xy = 12 hence solutions are z = 3-2i and z = -3+2i

    z^3 = 1+i = ..., √2 cis(π/4), √2 cis(9π/4), √2 cis(17π/4), ... {3 consecutive occurrences for the exponent of 3}

    so z = ⁶√2 cis(π/12), ⁶√2 cis(3π/4), ⁶√2 cis(17π/12)

    z^2 + (1+4i)z + (-3+3i) = 0 {by competing and factoring the difference of two squares - old school}

    z^2 + 2(½+2i)z + (½+2i)^2 + (-3+3i) - (½+2i)^2 = 0

    (z + ½ + 2i)^2 + (-3+3i) + (-¼ + 4 - 2i) = 0

    (z + ½ + 2i)^2 + (-3 - ¼ + 4) + (3-2)i = 0

    (z + ½ + 2i)^2 + (¾ + i) = 0

    (z + ½ + 2i)^2 - i^2(1+½i)^2 = 0

    (z + ½ + 2i)^2 - (-½ + i)^2 = 0

    (z + ½ + 2i + ½ - i) (z + ½ + 2i - ½ + i) = 0

    (z+1+i) (z+3i) = 0

    Solutions: z = -1-i and z = -3i

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