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Imaginary numbers multiply and divide

  1. Oct 26, 2011 #1
    I don't really understand how to multiply and divide when numbers are in a+bi form
  2. jcsd
  3. Oct 26, 2011 #2


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    The operations work exactly the same as for just Real Numbers. That's in the beginning. Later, you learn properties of dealing with complex numbers. They are developed or derived logically.

    You will learn what is a Conjugate of a complex number, and what happens if you multiply a complex number by its conjugate.

    I do not fill in much for you here. You can study about Complex Numbers both in a Trigonometry book and in a College Algebra book.
  4. Oct 26, 2011 #3
    A complex number has the form x + iy, where x is called the real part and y is the imaginary part.

    (a+ib)(c+id) = (ac - bd) + i(cb + ad)

    This looks weird, but just factor it out and you will see why it happens:

    (a+ib)(c+id) = ac + iad + ibc + i^2*bd

    but we know that i = sqrt(-1) so i^2 = -1 so this becomes

    ac + iad + ibc - bd = (ac - bd) + i(cb + ad)

    Dividing complex numbers is a little different. FIrst realize that if z = x+iy is a complex number, its complex conjugate is x - iy. We can imagine a graph:
    (+ imaginary axis)
    |-(iy) * (x+iy)
    ------|------- (+ x axis)
    |-(-iy) * (x-iy)
    (- imaginary axis)

    We also know that the modulus of a complex number z (its distance from itself to the origin) is sqrt(x^2 + y^2)

    If z = x + iy is a complex number, then 1/z = (x - iy)/(x^2 + y^2)

    So think about what this means, first realize that we can break this into something clearer:

    x/(x^2+y^2) - i(y/x^2+y^2) So first we are taking the complex conjugate, but first dividing the real and imaginary parts by the square of the modulus

    Realize we have the same familiar rules: associative, commutative, distributive...
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