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Homework Help: Complex numbers

  1. Apr 1, 2005 #1
    Please help with this question. I'm really having trouble.

    Let z= r(cosa+isina) be any complex number.

    a. Show that lz^2l = lzl^2 (I know how to prove this question, but I just put this down because I'm assuming you'd need it for the following questions)

    b. For which (if any) complex numbers is Re(z^2) = (Rez)^2

    c. For which (if any) complex numbers is Im(z^2) = (Imz)^2

    I tried solving b. and I ended up with an answer involving "cos".

    Any help would be greatly appreciated.
  2. jcsd
  3. Apr 1, 2005 #2


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    Do you mean you got that this would only be true if cos a was a certain value? (let me guess, 1 or -1) ? That's fine, just find out what this means about sin a (if you got what I guessed, sin a must be 0), plug it into z and see what you get. It should be pretty clear why this is the case.
  4. Apr 1, 2005 #3
    Try it by converting [itex]z[/itex] back to cartesian coordinates (ie. set [itex]z = a + bi \Longrightarrow z^2 = a^2 - b^2 + 2abi[/itex]).

    I'm sure if you think about it for a little while you can come up with a reasonable guess for part b without doing any calculation anyways :wink:
  5. Apr 4, 2005 #4
    How did you get [itex]z^2 = a^2 - b^2 + 2abi[/itex]? I must be blind or something. Sorry about this. I just don't understand how the negative sign becomes involved and the "i" disappears in b^2.

    I asked my tutor and he also said to convert it to cartesian form.
    Last edited: Apr 4, 2005
  6. Apr 4, 2005 #5
    if [itex]z= a+bi[/itex] then

    [tex]z^2 = (a+bi)(a+bi) = a^2 + 2abi + (bi)^2 = a^2 + 2abi + (-1)b^2 = a^2 - b^2 + 2abi.[/tex]
  7. Apr 4, 2005 #6
    Thanks I forgot i^2 equals -1. Sorry, I'm completely lost...do you find the real part of that equation for part a.?
  8. Apr 4, 2005 #7
    you mean part b, right?

    If so, then it's asking you to find out when

    [tex] \left(\mbox{Re} [z]\right)^2 = \mbox{Re} \left[z^2\right][/tex]

    so yes :smile:.
  9. Apr 4, 2005 #8
    Yeah, sorry, i meant part b. I can't seem to get it. The real part of that equation is a^2 - b^2, for Re(z)^2? But for (Re[z])^2, do you find the real z from a+ib (which is a)? I'm assuming I'm doing something wrong. We havent exactly been taught this stuff. :redface:
  10. Apr 4, 2005 #9
    Well, you correctly identified

    [tex]\mbox{Re}\left[z^2\right] = a^2 - b^2[/tex]


    [tex]\mbox{Re}[z] = a \Longrightarrow \left( \mbox{Re}[z]\right)^2 = a^2[/tex]

    so, can you tell me under what conditions

    [tex]a^2 = \left( \mbox{Re}[z]\right)^2 = \mbox{Re}\left[z^2\right] = a^2 - b^2[/tex]

  11. Apr 4, 2005 #10
    When b^2 = 0?
  12. Apr 4, 2005 #11
    Right. So if [itex]b^2=0 \Longleftrightarrow b = 0[/itex], then what is [itex]z[/itex]?
  13. Apr 4, 2005 #12
    a? (10 char)
  14. Apr 4, 2005 #13
    right... so for what complex numbers [itex]z[/itex] do we have the result part b wants?

    remember that [itex]a[/itex] is real!
  15. Apr 4, 2005 #14
    Is it any real number for a and 0 for b? Now, I'm not completely sure about this.
  16. Apr 4, 2005 #15
    Indeed. In other words,

    [tex]\left(\mbox{Re}[z]\right)^2 = \mbox{Re}\left[z^2\right][/tex]

    for [itex]z \in \mathbb{C}[/itex] if and only if [itex]z[/itex] is real, or in other words [itex]z \in \mathbb{R}[/itex] (ie. [itex]z[/itex] has no imaginary part!).

    Do you see why this would have been a reasonable guess from the start?

    As I'm sure you know, the real part of any real number is just the number itself, and the square of any real number is real also! :wink:
  17. Apr 4, 2005 #16
    Thank you so much :smile:

    So, for the imaginary part

    Im(2abi) = Im(i^2*b^2)

    Im (2abi) = Im (-b^2)

    So, I have to find out what condidtions Im (2abi) = Im (-b^2)?

    So, 2abi = -b^2

    How exactly do I go about solving this. Obviously 2ai = -b, but how do you get a real number from this?
  18. Apr 4, 2005 #17
    Remember, the "imaginary part" is just the real coefficient of [itex]i[/itex]. So if [itex]z = a + bi, \; (a, b \in \mathbb{R})[/itex] then [itex]\mbox{Im}[z][/itex] is just [itex]b[/itex].
  19. Apr 4, 2005 #18
    Ok...Im[2ab] = Im[b^2]

    2ab = b^2

    2a = b

    I'm not sure how to get a real number out of this.
  20. Apr 4, 2005 #19
    Both sides of that equation are real... and you aren't taking [itex]\mbox{Im}[2ab][/itex] or [itex]\mbox{Im}\left[b^2\right][/itex], either.

    Here's how it should be presented formally:

    Let [itex]z = r\left( \cos \theta + i \sin \theta \right) = a + bi, \ (a, b \in \mathbb{R})[/itex].

    Then [itex]z^2 = a^2 - b^2 + 2abi[/itex] so

    [tex]\mbox{Im}\left[z^2\right] = 2ab[/tex]

    NOTE: here I take the imaginary part of [itex]z^2[/itex]. This function returns to me the real coefficient of [itex]i[/itex] when [itex]z^2[/itex] is expressed in standard form, or in this case, [itex]2ab[/itex]. On the other hand, writing it the way you did, really, [itex]\mbox{Im}[2ab] = 0[/itex] since [itex]2ab[/itex] has no imaginary part!

    Continuing, we now look at [itex]\mbox{Im}[z]\right = b \Longrightarrow \left(\mbox{Im}[z]\right)^2 = b^2[/itex].

    Thus we find that [itex]\mbox{Im}\left[z^2\right] = \left(\mbox{Im}[z]\right)^2[/itex] precisely whenever

    [tex]2ab = b^2 \Longrightarrow 2a = b \ \mbox{or} \ b=0[/itex]

    so we find that the result holds whenever [itex]z[/itex] has the form [itex]z = [/itex]_______________.

    Fill in the blank! :smile:
    Last edited: Apr 4, 2005
  21. Apr 4, 2005 #20
    a + 2ai (do I include the imaginary part)?
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