Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Dividing imaginary numbers... a+bi?

4-3i / 5+5i

How do you do that?

4 Answers

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  • 1 decade ago
    Favorite Answer

    4-3i / 5+5i

    You need to make the denominator real, so multiply the entire thing by 5-5i/5-5i to get:

    (4-3i)(5-5i)/(5+5i)(5-5i)

    This foils out to:

    20-20i-15i+15i^2/25-25i+25i-25i^2

    This reduces to:

    5-35i/50

    You can reduce this to:

    1-7i/25

    To put it into standard form:

    1/25 - 7/25i

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  • 1 decade ago

    multiply by conjugate/conjugate of the denominator.

    conjugate of a +bi is a - bi

    so multiply 4-3i/5+5i by 5-5i/5-5i

    the denominator will have no i terms because i^2 = -1

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  • mammo
    Lv 4
    3 years ago

    rationalize 7i(-9 +14i) / 7i(7i) (-63i+ 98i^2) / 49i^2 be conscious that i^2 = -a million (-63i +ninety 8(-a million)) / 40 9(-a million) 7 (-9i -14) / -40 9 be conscious that 7 is going into 7 as quickly as and into 40 9 seven circumstances -(9i +14) / -7 9i/7 + 14/7 9i/7 + 2 2 + 9i/7

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  • Como
    Lv 7
    1 decade ago

    (4 - 3i)(5 - 5i)

    ------------------

    (5 + 5i)(5 - 5i)

    20 - 20 i - 15 i + 15 i ²

    -----------------------------

    25 - 25 i ²

    5 - 35 i

    --------------

    25 + 25

    5 - 35 i

    -----------

    50

    1 - 7 i

    ---------

    10

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