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# Parametic equation of a line in complex variable. watch

1. I am having trouble figuring out the line segment between and where t is between .

The normal equation for a line, i.e, doesn't appear to work and I'm unsure how to do it.
2. Maybe I'm missing something, but I'm pretty sure it does work.
3. I thought so too before and worked it out that way and got however according to the answer sheet the answer should be and I don't see why. Apparently the fact that t is not between and is between makes a difference. I'm really confused so I don't know how credible the statements are. These are just assumptions because I am not getting the answer using that line equation.
4. It is 1 +i -2it for 0<=t<=1 using definition of line segment.

Write 1+i-2it = 1 + i(1-2t), and put u=1-2t, then the line segment is equivalent to

1+iu, for -1<=u<=1.
5. If t goes between -1 and 1, and your line goes between 1-i and 1+i, notice that the real part is fixed and the imaginary part goes between -1 and 1... so it would make sense for t to be precisely the imaginary part of the line. That is, . Plugging it into doesn't work because that's for when t goes from 0 to 1, not -1 to 1. The corresponding formula for when t goes from -1 to 1 would be , but that seems to complicate things here to be honest.
6. (Original post by nuodai)
If t goes between -1 and 1, and your line goes between 1-i and 1+i, notice that the real part is fixed and the imaginary part goes between -1 and 1... so it would make sense for t to be precisely the imaginary part of the line. That is, . Plugging it into doesn't work because that's for when t goes from 0 to 1, not -1 to 1. The corresponding formula for when t goes from -1 to 1 would be , but that seems to complicate things here to be honest.
When you say the equation would correspond to the imaginary part of the line and decude that it is, ... I'm just unable to link those two statements. Could you please run over that part again?

And normally the equation of a circle is defined to be for when t is between so similarly if the range for t is between , how do things change?
7. (Original post by canine101)
When you say the equation would correspond to the imaginary part of the line and decude that it is, ... I'm just unable to link those two statements. Could you please run over that part again?
Your parameter is running from -1 to 1, and your line is running from 1-i to 1+i. You'll notice that the imaginary part of 1-i is -1 and the imaginary part of 1+i is 1. Since this is a straight line and the real part of 1-i and 1+i is the same in both cases (namely, 1), the only thing that changes along the line is the imaginary part, and the imaginary part changes from -1 to 1. Because the parameter also changes from -1 to 1, it would make sense for the imaginary part of a point on the line to be the parameter value. That is, if lies on the line, then , and this runs from -1 to 1. Because we know that everywhere along the line, and because we know that , we just put these values in to get .

(Original post by canine101)
And normally the equation of a circle is defined to be for when t is between so similarly if the range for t is between , how do things change?
No, the equation of a circle is given by for . If you only let t go between 0 and 1 then you'd only get a circular arc (which subtends an angle 1 radian). Similarly if you let t go between -1 and 1 then you'd get another circular arc, which subtends an angle of 2 radians.

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Updated: April 5, 2011
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