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Math Help - Can someone explain to me how you execute "Re" and "Im" in a formula?

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    Can someone explain to me how you execute "Re" and "Im" in a formula?

    I know that in a complex number such as

    z=a+bi

    "Re" is supposed to be the "real part", which is a so that Re(a+bi)=a and that "Im" is supposed to be the "imaginary part", which is b so that Im(a+bi)=b in the above example.

    However, recently I stumpled upon a problem when trying to solve linear systems:

    y(t)=Re(H(i\omega)e^{i\omega t})

    I'm not sure that I understand how exactly to execute this formula with the "Re" part.

    Here is an example:
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)-\frac{4}{25}\sin(t)

    Can somebody explain to me what the "Re" did here?

    And what about "Im"? If it had said "Im" instead, what would be different in the answer?
    Last edited by MathIsOhSoHard; September 22nd 2012 at 11:50 AM.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Edit: when I look at it more closely Im confused myself... is t a real variable?
    Last edited by SworD; September 22nd 2012 at 12:11 PM.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by MathIsOhSoHard View Post
    I know that in a complex number such as

    z=a+bi

    "Re" is supposed to be the "real part", which is a so that Re(a+bi)=a and that "Im" is supposed to be the "imaginary part", which is b so that Im(a+bi)=b in the above example.
    However, recently I stumpled upon a problem when trying to solve linear systems:
    y(t)=Re(H(i\omega)e^{i\omega t})
    I'm not sure that I understand how exactly to execute this formula with the "Re" part.
    Here is an example:
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)-\frac{4}{25}\sin(t)
    Your example is incorrect.
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)
    AND
    Im\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{4}{25}\cos(t)+\sin(t)

    Now did you copy it correctly.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    in general, for a complex number z:

    \operatorname{Re}(z) = \frac{z + \overline{z}}{2}

    \operatorname{Im}(z) = \frac{z - \overline{z}}{2i}

    so if z = a+bi, then \operatorname{Re}(z) = a, \operatorname{Im}(z) = b.

    for example:

    \operatorname{Re}(e^{i\omega t}) = \cos(\omega t).

    what you want to do, if you have complex expressions multiplied together, is first "multiply them out", to get something in the form a+bi.

    EDIT: i suspect what you actually WANT is:

    \operatorname{Re}\left(\left(\frac{-3+4i}{25}\right)(\cos(t) + i\sin(t))\right)

    parentheses matter!!!!

    if so then the real part is:

    \frac{-3}{25}\cos(t) - \frac{4}{25}\sin(t)
    Last edited by Deveno; September 22nd 2012 at 12:51 PM.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by Plato View Post
    Your example is incorrect.
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)
    AND
    Im\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{4}{25}\cos(t)+\sin(t)

    Now did you copy it correctly.
    Yes, I have looked at it 5 times now and my book writes this as a solution. Here it is in its entirety:

    A linear system is given as:
    \frac{dx}{dt}=\begin{bmatrix}2 & -1 \\ 3 & 1 \end{bmatrix}x(t)+\binom{0}{-1}u(t)
    And y(t)=2x_1(t)-x_2(t)

    The transfer function is calculated to H(s)=\frac{s}{5-3s+s^2} and given that u(t)=cos(t) then a linear system has the solution:

    y(t)=Re(H(i\omega)e^{i\omega t})

    Where u(t)=cos(\omega t)

    Since in this example, \omega=1 we have the formula for the solution:

    y(t)=Re(H(i)e^{it})

    Replacing "s" with the imaginary unit in the transfer function:

    H(i)=\frac{i}{5-3i+i^2}

    Inserting into the solution gives:

    y(t)=Re\left( \frac{i}{5-3i+i^2}e^{it} \right)

    Using Euler's formula we now have:

    y(t)=Re\left( \frac{i}{5-3i+i^2}\cos(t) + i \sin(t) \right)

    Which according to my book:
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)-\frac{4}{25}\sin(t)
    Last edited by MathIsOhSoHard; September 22nd 2012 at 12:57 PM.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Yes, t is a real variable.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    see my post above. you are missing parentheses. people are thinking you mean the fraction is just times the cosine term, instead of the entire cosine + imaginary sine expression.

    what you have are two complex numbers multiplied together.

    now (a+bi)(c+di) = (ac - bd) + (ad + bc)i, and the real part of this is ac - bd.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by Deveno View Post
    see my post above. you are missing parentheses. people are thinking you mean the fraction is just times the cosine term, instead of the entire cosine + imaginary sine expression.

    what you have are two complex numbers multiplied together.

    now (a+bi)(c+di) = (ac - bd) + (ad + bc)i, and the real part of this is ac - bd.
    Ah yes, that is true. This is how it should be then:

    Re\left( \frac{-3+4i}{25}(\cos(t)+i\cdot\sin(t)) \right)

    so that leaves me with:

    Re\left( \frac{-3+4i}{25}\cos(t)+\frac{-3+4i}{25}i\cdot\sin(t)) \right)=-\frac{3}{25}\cos(t)+???

    But isn't 4i a part of the imaginary part? So how does that fit into the solution? :S
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by MathIsOhSoHard View Post
    Yes, I have looked at it 5 times now and my book writes this as a solution. Here it is in its entirety:

    Inserting into the solution gives:

    y(t)=Re\left( \frac{i}{5-3i+i^2}e^{it} \right)

    Using Euler's formula we now have:

    y(t)=Re\left( \frac{i}{5-3i+i^2}\cos(t) + i \sin(t) \right)

    Which according to my book:\
    Re\left( \frac{-3+4i}{25}\cos(t)+i\cdot\sin(t) \right)=\frac{-3}{25}\cos(t)-\frac{4}{25}\sin(t)
    I see several mistakes.
    Re\left( \frac{i}{5-3i+i^2}e^{it} \right) should be written as y(t)=Re\left( \frac{i}{5-3i+i^2}\left(\cos(t)+i\sin(t) \right) \right) There is a set of () missing.

    \text{Re}\left[(a+bi)(c+di)=ac-bd. If you multiply correctly text is correct if you use the extra set of ().
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by Plato View Post
    I see several mistakes.
    Re\left( \frac{i}{5-3i+i^2}e^{it} \right) should be written as y(t)=Re\left( \frac{i}{5-3i+i^2}\left(\cos(t)+i\sin(t) \right) \right) There is a set of () missing.

    \text{Re}\left[(a+bi)(c+di)=ac-bd. If you multiply correctly text is correct if you use the extra set of ().
    Yes that is true about the parentheses. But I'm still not sure I understand the last part.

    So we have that [TEX]\frac{-3+4i}{25}[TEX] where the 4i is the imaginary part, so that disappears, leaving:
    -\frac{3}{25}\cos(t)

    But then I don't understand where the fraction \frac{4}{25} comes from. Isn't 4i the imaginary part?

    What is the multiplication rule that you mention? I haven't seen that before.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    I just used the Re(a+bi)(c+di)=ac-bd and that worked out really well!
    But I've never seen this rule before? :S
    Also is there a similar rule for Im?

    Okay, is this correct then?

    Re((a+bi)(c+di))=ac-bd
    Im((a+bi)(c+di))=bci+adi

    I think I got it now?
    Last edited by MathIsOhSoHard; September 22nd 2012 at 01:27 PM.
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by MathIsOhSoHard View Post
    Yes that is true about the parentheses. But I'm still not sure I understand the last part.

    So we have that [TEX]\frac{-3+4i}{25}[TEX] where the 4i is the imaginary part, so that disappears, leaving:
    -\frac{3}{25}\cos(t)

    But then I don't understand where the fraction \frac{4}{25} comes from. Isn't 4i the imaginary part?

    What is the multiplication rule that you mention? I haven't seen that before.
    \left( \frac{-3+4i}{25}\left(\cos(t)+i\sin(t) \right) \right)=\left[ {\frac{{ - 3\cos (t)}}{{25}} - \frac{{4\sin (t)}}{{25}}} \right] + i\left[ {\frac{{ - 3\sin (t)}}{{25}} + \frac{{4\cos (t)}}{{25}}} \right]
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    it's very simple.

    to find the real part of a complex number easily, you have to have it in the form:

    (something real) + i (something else that's real).

    now what you have is a complex fraction (a+bi)/d. that is the complex number (a/d) + (b/d)i.

    you have it multiplied by another complex number: cos(t) + i sin(t).

    when we multiply these together:

    [(a+bi)/d](cos(t) + i sin(t)), we get:

    (a/d)cos(t) - (b/d)sin(t) + [(a/d)sin(t) + (b/d)cos(t)]i

    the real part is thus: (a/d)cos(t) - (b/d)sin(t) and the imaginary part is:

    (a/d)sin(t) + (b/d)cos(t)

    in your example above:

    a = -3
    b = 4
    d = 25

    yes, the imaginary part of (a+bi)(c+di) is: ad+bc. why? this is the very DEFINITION of complex multiplication:

    (a+bi)(c+di) = a(c+di) + (bi)(c+di) = ac + (ad)i + (bi)c + (bi)(di) = ac + (ad)i + (bc)i + (bd)i2.

    but i2 = -1, so:

    ac + (ad)i + (bc)i + (bd)i2 = ac + (ad)i + (bc)i + bd(-1) = ac - bd + (ad + bc)i (collecting the "i" terms).
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Thanks both of you!

    Just to see if I got it, if I in my example had "Im" instead of "Re" would this then be correct?

    Im\left( \frac{-3+4i}{25}(\cos(t)+i\cdot\sin(t)) \right)=Im\left( \left( \frac{-3}{25} + \frac{4}{25} i \right) \left( \cos(t) + i\cdot \sin(t)\right) \right)=\frac{4}{25}\cdot i \cos(t) + \frac{-3}{25}\cdot i \sin(t)
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    Re: Can someone explain to me how you execute "Re" and "Im" in a formula?

    Quote Originally Posted by MathIsOhSoHard View Post
    Just to see if I got it, if I in my example had "Im" instead of "Re" would this then be correct?
    Im\left( \frac{-3+4i}{25}(\cos(t)+i\cdot\sin(t)) \right)=Im\left( \left( \frac{-3}{25} + \frac{4}{25} i \right) \left( \cos(t) + i\cdot \sin(t)\right) \right)=\frac{4}{25}\cdot i \cos(t) + \frac{-3}{25}\cdot i \sin(t)
    There is no i in the imaginary part.

    \text{Im}(a+bi)=b NOT bi. I would mark that wrong.
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