Locus on argand diadram. Watch

Tallon
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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.
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SimonM
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Ermm... z is the variable so it should be a circle, centre a

arg(z-a) is the angle between \overline{AZ} and the origin. You need to make that angle \frac{5 \pi}{8}. Therefore it will be a half line (I'll leave you to figure out where and at what angle)

Also is a = 2 ( \cos \frac{\pi}{8} + i \sin \frac{\pi}{8} )?
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Tallon
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yeah, it is
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Tallon
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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?
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SimonM
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(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?
|z-a|=6

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
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Mr M
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Tallon,

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

z=x+yi

a=p+qi

|z-a|=6

|(x-p)+(y-q)i|=6

\sqrt{(x-p)^2+(y-q)^2}=6

(x-p)^2+(y-q)^2=6^2

This is the circle centre a radius 6.
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SimonM
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Mr M, a is not a real number. He made a mistake in the first post
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Mr M
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(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.
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Tallon
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I know the second part is a completely seperate question. that's the bit I'm stuck on lol
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Mr M
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(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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