i dont know how to solve this. can someone tell me step by step without giving away the answer... (2+8i)(2-8i) thank you in advance...?

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  • 3 months ago

    (2 + 8i)(2 - 8i)

    Difference of Squares, or use FOIL

    = 2² – (8i)²

    = 2² – 64 * i²

    i² = -1

    = 2² – 64(-1)

    = 4 +64

    =

  • 3 months ago

    (2 + 8i)(2 - 8i)

    = 2² – (8i)²

    = 4 + 64

    = 68

  • Amy
    Lv 7
    3 months ago

    Remember FOIL? If not, just apply the distributive property twice:

    (a + b)(c + d) = (a+b)c + (a+b)d = ac + bc + ad + bd

    When you apply this to the complex numbers, you will get a combination of real and imaginary terms. Remember that i^2 = -1, so the term with i^2 becomes a real number. Add the real terms together, and add the imaginary terms together.

  • sepia
    Lv 7
    3 months ago

    (2 + 8i)(2 - 8i)

    = 68

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  • 3 months ago

    Expanding gives:

    4 - 16i + 16i - 64i²

    With i² = -1 we have:

    4 - 16i + 16i - 64(-1)

    i.e. 4 - 16i + 16i + 64

    You can complete this?

    :)>

  • 3 months ago

    (2+8i)(2-8i)

    =

    2^2-(8i)^2

    =

    4+64

    =

    68

    [Note that (a+b)(a-b)=a^2-b^2;

    Now a=2, b=8i. Also, (8i)^2=

    (8i)(8i)=64i^2= -64 because

    by definition i^2=-1]

  • Mike G
    Lv 7
    3 months ago

    2^2-64i^2

    *******************

  • TomV
    Lv 7
    3 months ago

    Multiply each of the terms within the second set of parentheses by each term within the first set of parentheses and combine like terms of the result.

    For a general example consider (a+ib)(a-ib)

    (a+ib)(a-ib) = a(a-ib) + ib(a-ib)

    = a(a) - a(ib) + ib(a) - ib(ib)

    = a² - aib + aib - i²b²

    = a² - i²b²

    Now consider that if i = √-1, then i² = -1, and the summation becomes:

    = a² - (-1)b²

    = a² + b²

    Apply that general process to your specific problem to obtain the answer to (2+8i)(2-8i)

    (Hint: let a = 2 and b = 8)

  • 3 months ago

    (2+8i)(2-8i)

    I don't see how I can solve it without giving you the answer...

    long way

    2(2-8i) + 8i(2-8i)

    4 – 16i + 16i – 64i²

    note: 1² = –1

    4 – 64(–1)

    4 + 64

    68

    fast way, using identity a² – b² = (a + b)(a – b)

    (2+8i)(2-8i)

    4 – (8i)²

    4 + 64 = 68

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