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Homework Help: Complex numbers. Imaginary part

  1. Jan 14, 2004 #1
    z1 = x + iy
    z2 = x - iy
    (Complex conjugate)

    Im (1/z1)

    This is what I have tried to do:

    (1) z1*z2 = x^2 + y^2

    (2) z2 / (x^2 + y^2) = 1 / z1

    The answer is:
    -y / (x^2 + y^2) = Im (1 / z1)

    So my question is:
    Can I change z2 to Im (z2) and z1 to Im (z1) in equation (2)?
    Last edited: Jan 14, 2004
  2. jcsd
  3. Jan 14, 2004 #2
    They want the coefficient of i or Im(1/z)

    i.e [tex]\frac{1}{x+iy}[/tex]

    What was/are ur thoughts???
  4. Jan 14, 2004 #3
    oops, read another post and found that it would be wise to post my work. And i reckoned that no-one would have had the time to reply so i just edited my post.

    But do my new post clarify anything?
  5. Jan 14, 2004 #4
    No u can't change them that ways. It will defy all the properties of complex number
  6. Jan 14, 2004 #5
    I think your work is correct. Because, if 2 numbers are equal, then their imaginary parts are equal.
  7. Jan 14, 2004 #6


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    Gold Member

    Yes, fannemel, you are entirely correct. Given z = x + iy, Im(1/z) is indeed -y / (x^2 + y^2).

    - Warren
  8. Jan 15, 2004 #7
    His Questions is
    [tex]\frac{Im_{z2}}{x^2 + y^2} = \frac{1}{Im_{z1}}[/tex]
    Which is not true
    Last edited: Jan 15, 2004
  9. Jan 15, 2004 #8
    what about
    [tex]\frac{Im_{z2}}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]

    Would that be any better?

    For me that would equal:

    [tex]{\frac{-y}{x^2 + y^2} = Im [\frac{1}{z1}][/tex]
    since [tex]Im_{z2} = -y [/tex]
    Last edited: Jan 15, 2004
  10. Jan 17, 2004 #9
    Yes that is absolutely correct. You get the imaginary part of the reciprocal of z1, not z1
  11. Jan 17, 2004 #10


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    That is true because [tex]x^2+ y^2[/tex] is a real number.
    In general you cannot get the imaginary part of a number computed by a formula just by replacing each number in the formula by its imaginary part.

    It would be far better for you to replace [tex]z_1[/tex] and [tex]z_2[/tex] by x+iy and x-iy right from the start:

    [tex]\frac{1}{z_1}= \frac{1}{x+iy}[/tex]. Now multiply both numerator and denominator by x- iy to get [tex]\frac{(1)(x- iy)}{(x+iy)(x-iy)}= \frac{x- iy}{x^2+ y^2}= \(\frac{x}{x^2+y^2}\)-\(\frac{y}{x^2+y^2}\)i[/tex] so that it is obvious that the real part is [tex]\frac{x}{x^2+y^2}[/tex] and the imaginary part is [tex]\frac{-y}{x^2+y^2}[/tex].
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