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Homework Help: Identify the imaginary part

  1. Jan 31, 2006 #1
    I am supposed to identify the imaginary part (marked in bold) of each expression, just wanted to see if I got them correct:

    1. (1+i)+(1-i) ..........................0
    2. (5+i)+(1+5i) ..........................6
    3. (5+i)-(1-5i) ..........................6
    4. 1+2i+3+4i+5 ..........................6
    5. [tex]S=1+\frac{1}{2i}+\frac{1}{4}+\frac{1}{8i}+\frac{1}{16}+\frac{1}{32i}+\frac{1}{64}+...[/tex]

    using the geometric series: (.5)/(1-.25)=2/3=1/1.5i
    does this mean the imaginary part is 1.5 since the i is in the denominator? or is it 1.5?

    6. a and b are constants
    a+bi ..........................b
    7. (a+ib)^2 ..........................2ab
    8. (a+bi)(b+ia) ..........................a^2+b^2
    9. (ia)^3 ..........................-a^3
    10. a(a+i)(a+2i) ..........................3a^2

    are these correct?
     
    Last edited: Jan 31, 2006
  2. jcsd
  3. Jan 31, 2006 #2

    VietDao29

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    The way I see it, everything is correct except the 5th one
    If the complex number is in the form: a + ib then a is the real part, and b is the imaginary part. You must do something to make it have the form a + ib, i.e i cannot be in the denominator, it must be in the numerator.
    Hint:
    [tex]\frac{a}{ib} = \frac{ia}{i ^ 2b} = -\frac{ia}{b}[/tex]
    Multiply both numerator and denominator by i. :smile:
     
  4. Jan 31, 2006 #3
    then it is -1/15?
     
  5. Jan 31, 2006 #4

    VietDao29

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    Nope, you should check your answer again.
    The first term is -1 / 2, and the common ratio is still 1 / 4.
    It should be -2 / 3. Shouldn't it? :smile:
     
  6. Jan 31, 2006 #5

    HallsofIvy

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    The imaginary part of
    [tex]S=1+\frac{1}{2i}+\frac{1}{4}+\frac{1}{8i}+\frac{1} {16}+\frac{1}{32i}+\frac{1}{64}+...[/tex]
    is
    [tex]S=\frac{1}{2i}+\frac{1}{8i}+\frac{1}{32i}+...[/tex]
    which is indeed, a geometric sequence,
    [tex]S= \frac{1}{2i}\left(1+ \frac{1}{4}+ \frac{1}{16}+...\right)[/tex]
    having sum
    [tex]\frac{1}{2i}\frac{1}{1- \frac{1}{4}}[/tex]
    [tex]= \frac{1}{2i}\frac{4}{3}= \frac{2}{3}\frac{1}{i}[/tex]
    which is what you got. The only question is "what do you do with that 1/i?"

    Well, i*i= -1, of course, so i(-i)= -(-1)= 1. -i is the multiplicative inverse of i: 1/i= -i.
    [tex]\frac{2}{3}\frac{1}{i}= -\frac{2}{3}i[/tex].
     
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