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Math Help - Complex Analysis (A Few Clarifying Questions On Practice Problems)

  1. #1
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    Complex Analysis (A Few Clarifying Questions On Practice Problems)

    I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

    1) Find a parameterization of the line segment joining and . Find the image of the segment under the mapping .
    This is what I did:
    and
    for
    =2i(1-t)+(1+t)i
    =2i-2t+i-1
    =3i-2it-1
    f(z)=i(\bar{3i-2it-1})
    =i(\bar{-1+3-2t)i})
    =i(-1-(3-2t)i)
    =i(-1-(3i-2it))
    =i(-1-3i+2it)
    =-i-3i^{2}+2i^{2}t
    =-i+3-2t for 0\leq t \leq 1

    The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

    2)Find the image S' of the set S under the given complex mapping w=f(z)
    f(z)=\bar{z}; S is the line y=x

    A friend told me to do it this way:
    Re(z)=Im(z) \therefore w=\bar{z}=Re(z)=-Im(z)
    Is this correct?

    I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.


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  2. #2
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    Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

    Quote Originally Posted by DrKittenPaws View Post
    I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

    1) Find a parameterization of the line segment joining and . Find the image of the segment under the mapping .
    This is what I did:
    and
    for
    =2i(1-t)+(1+t)i
    =2i-2t+i-1
    =3i-2it-1
    f(z)=i(\bar{3i-2it-1})
    =i(\bar{-1+3-2t)i})
    =i(-1-(3-2t)i)
    =i(-1-(3i-2it))
    =i(-1-3i+2it)
    =-i-3i^{2}+2i^{2}t
    =-i+3-2t for 0\leq t \leq 1

    The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

    2)Find the image S' of the set S under the given complex mapping w=f(z)
    f(z)=\bar{z}; S is the line y=x

    A friend told me to do it this way:
    Re(z)=Im(z) \therefore w=\bar{z}=Re(z)=-Im(z)
    Is this correct?

    I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.


    the parameterization of the segment is simply

    $2\imath + (1-\imath)t~~0\leq t \leq 1$

    try redoing the problem with this.
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  3. #3
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    Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

    Quote Originally Posted by romsek View Post
    the parameterization of the segment is simply

    $2\imath + (1-\imath)t~~0\leq t \leq 1$

    try redoing the problem with this.
    Well, that was dumb of me. Let me try to work it out again and I'll get back to ya on what I got. Thanks!
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  4. #4
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    Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

    Quote Originally Posted by DrKittenPaws View Post
    I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.

    1) Find a parameterization of the line segment joining and . Find the image of the segment under the mapping .
    This is what I did:
    and
    for
    =2i(1-t)+(1+t)i
    =2i-2t+i-1
    =3i-2it-1
    f(z)=i(\bar{3i-2it-1})
    =i(\bar{-1+3-2t)i})
    =i(-1-(3-2t)i)
    =i(-1-(3i-2it))
    =i(-1-3i+2it)
    =-i-3i^{2}+2i^{2}t
    =-i+3-2t for 0\leq t \leq 1

    The graph doesn't look look right. I've reworked it a few times and I keep getting the same answer. Am I making a silly algebraic mistake somewhere?

    2)Find the image S' of the set S under the given complex mapping w=f(z)
    f(z)=\bar{z}; S is the line y=x

    A friend told me to do it this way:
    Re(z)=Im(z) \therefore w=\bar{z}=Re(z)=-Im(z)
    Is this correct?

    I'll just submit the two problems I'm having the most problem with. Thanks for looking at this.


    You should have $z_0(1 - t)$ and not $z_0(1 + t)$.

    For your second problem, you have the elements on your line are complex numbers of the form: $z = x + ix$.

    From this, we see that the images are of the form: $f(z) = \overline{z} = \overline{x+ix} = x - ix$, so that these lie on the line $y = -x$.
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  5. #5
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    Re: Complex Analysis (A Few Clarifying Questions On Practice Problems)

    Quote Originally Posted by DrKittenPaws View Post
    I am having a few issues with Complex Analysis. I've done as much as I could on a few practice problems hoping I would understand it but I'm a bit stuck.
    1) Find a parameterization of the line segment joining and . Find the image of the segment under the mapping .
    Consider a vector approach. Express the points a vectors.
    $(0,2)~\&~(1,1)\\(0,2)+t(1,-1),~0\le t\le 1\\(t,2-t),~0\le t\le 1\\t+i(2-t),~0\le t\le 1$
    that is the parameterzation of the line segment.
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