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    ri0t's Avatar
    ri0t Posts: 13, Reputation: 2
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    #1

    Oct 21, 2009, 05:02 AM
    Complex Number - Locus
    Find the locus of the following: arg(z-3) - arg(z+2) = pi/6

    So far THis is what I have done:

    1. I have plotted the two points (on the complex plane)---> (-2,0) and (3,0)

    2. Now the locus would go in a circle, with these two points on the circumference.

    3. Then I put the angle that is subtended by the chord of these two ponts at the circumference as pi/6

    4. The centre of that circle would be (0.5, 0)

    5. Hence the angle subtended by the chord at the centre would be pi/3 (angle at centre is twice than angle at circumference)

    That is what I got up to. I know my explanation might be hard to visualise So I tried to draw up what I had in word and is attached, please have a look.

    Thank You

    ri0t
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  1. File Type: pdf Complex Number Question - AskMeHelpDesk - ri0t.pdf (151.9 KB, 226 views)
  2. Nhatkiem's Avatar
    Nhatkiem Posts: 120, Reputation: 9
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    #2

    Oct 21, 2009, 01:09 PM

    Is your drawing suppose to be to scale? Because if it is, those angles are definitely wrong.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Oct 21, 2009, 02:00 PM

    Hello ri0t. In step 4 - why do you say the center of the circle is at (0.5, 0)? Looks to me like the center is at (0.5, sqrt(3)).

    Also - in your drawing you show a full circle, including negatve y values, but I think it's really symmetric about the x axis - in other words, it looks like two partially overlapping circles.
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    ri0t's Avatar
    ri0t Posts: 13, Reputation: 2
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    #4

    Oct 22, 2009, 02:49 AM

    Ok, thank you,

    soz my step 4 was a mistake, I knew it wasn't at (0.5,0)
    I meant for it to say (0.5, y)

    so what will be description of the locus?

    also I don't get why it reflected about the x-axis?

    I don't think the answer can be blow the x-axis as the question says it = pi/6, hence eliminating any negative angles (in complex analysis, the domain of the arg is -pi < theta < pi.

    hence if it was below the x-axis it would read : arg(z-3) - arg (z+2) = - pi/6

    PS. My diagram was NOT to scale
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Oct 22, 2009, 05:50 AM

    It's been a long time (>30 years!) since I studied complex numbers in grad school, so I am a bit rusty at this. I see your point about the negative y domain - so the locus is just is the portion of the circle with positive values of y.

    Do you see why the radius of the circle is 2 and the center is at ?
    Chris-infj's Avatar
    Chris-infj Posts: 31, Reputation: 4
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    #6

    Oct 22, 2009, 09:59 PM


    ER, I beg to differ. Since the angle subtended by the (invisible) arc AB at the circumference is pi/6 then the angle at the centre of the circle will be pi/3. This means that the radius of the circle will be 5 and not 2. Then, the centre would have coordinates (0.5,y ) as stated in a previous post but then

    y = 5 cos (pi/6). That is centre (0.5, 4.33)

    And the locus of the complex number z will be anywhere on the edge of c excluding the points A and B.
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    ri0t Posts: 13, Reputation: 2
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    #7

    Oct 22, 2009, 10:24 PM

    Thanks a lot

    That really helped.

    BTW. What program did u use for that?
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    Chris-infj Posts: 31, Reputation: 4
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    #8

    Oct 22, 2009, 11:10 PM

    I used the free geometry package GeoGebra. See GeoGebra
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #9

    Oct 23, 2009, 05:53 AM
    Quote Originally Posted by Chris-infj View Post
    ER, i beg to differ. ... the radius of the circle will be 5 and not 2.
    Yes - my mistake. I'm aftraid that when I was working this I used cordinates (-1,0) and (1,0) as the real axis intercepts instead of (-2,0) and (3,0) to work through the math (it seemed easier that way) - which means my circle is 2/5 the size of the one as stated in the problem, and I forgot to convert back to the dimensions as stated. So yes - radius of 5.

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