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# Locus on argand diadram. watch

1. The complex number: a is given that:
a = 2((cos(pi/8) +sin(pi/8))

on your diagram, draw and label:
(z - a) = 6

the locus of points:
arg(z - a) = (5pi)/8

the first locus is just a circle, centre at z, with radius 6, right?

Need help, espeically with the last bit. Is the last question a shaded regiion in and around the line from the origin to complex number a that stretches on forever? which is 5pi/6 distance away from the line than goes trough a and onwards, and since it's 5pi/8 distance away from every point on that infinitely long line it's really just a shaded region? if that made sense. help please lol.
2. Ermm... z is the variable so it should be a circle, centre a

arg(z-a) is the angle between and the origin. You need to make that angle . Therefore it will be a half line (I'll leave you to figure out where and at what angle)

Also is ?
3. yeah, it is
4. Still a bit confused. Is it, centre A, a circle that goes from arga to 5pi/8, so it's not a full circle, but one that goes to that angle then stops?
5. (Original post by Tallon)
Still a bit confused. Is it, centre A, a circle that goes from arga to 5pi/8, so it's not a full circle, but one that goes to that angle then stops?

The locus of z is a circle centre a radius 6.

This is because z-a is the line from a to z and the modulus signs indicate that the length of this line is 6.

The second part is a completely separate question
6. Tallon,

If you want to understand the circle centre, consider this:

This is the circle centre radius .
7. Mr M, a is not a real number. He made a mistake in the first post
8. (Original post by SimonM)
Mr M, a is not a real number. He made a mistake in the first post
I wondered why he was finding it difficult.

My earlier post is now corrected in the light of this information.
9. I know the second part is a completely seperate question. that's the bit I'm stuck on lol
10. (Original post by Tallon)
I know the second part is a completely seperate question. that's the bit I'm stuck on lol
Private message me with an email address and I will send you a Powerpoint that will illuminate the matter for you.

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Updated: November 16, 2008
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