If K is positive and [z]</=K, prove that 0</=[z+K]</=2k and pi/2<arg(z+k),p/2
Find the least and greatest possible values of [z+2k] and arg[(z+2k)
That's the entire question.
Am I right in thinking that it's an easy proof with an argand diagram?
So the modulus of z is less than or equal to K, so I just draw a circle around the origin of radius K to start off with, right? And Z can be anything inside this cricle, yeah?
Next I draw a circle of the same radius, only shifted left K units along the real axis? The second circle goes as far as 2k left from the origon or on the origin exactly, right?
Then the part of the question about the arguments confuses me and makes me think I'm totally wrong altogether. Surely it would mean that arg(z+K) has to be a point to the right of the imaginary axis and I can't see how that would work. :s
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Tallon
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 16072009 19:00
Last edited by Tallon; 16072009 at 19:07. 
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 16072009 19:08
by "[....]" what do you mean?
Sorry im going to have a go at it but I don't know what everything means (I want to try and learn) 
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 16072009 19:15
Yeah, your notation is a bit strange. Generally people use z to mean the modulus of z and for lessthanorequalto people generally use =< or <=.
Your bit about the circles is right. For the arguments, note that K is positive, so draw z+k and think about what arg(z+k) means. It should be clearer. 
Tallon
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 16072009 20:35
(Original post by around)
Yeah, your notation is a bit strange. Generally people use z to mean the modulus of z and for lessthanorequalto people generally use =< or <=.
Your bit about the circles is right. For the arguments, note that K is positive, so draw z+k and think about what arg(z+k) means. It should be clearer.
Note that k is positive?
If I drew z+k, wouldn't it be a circle of a radius that could be a mininum of size k (if z is 0) and a max of size 2k (if z = k) centred around (k,0)? arg(z+k) is the angle from the point (k,0) isn't it? The angle all the points of z+k can be from that point? oh ffs i dont know.
That's what I don't understand. How could the argument possibly be between pi/2 and pi/2 if that's the case?Last edited by Tallon; 16072009 at 20:48. 
generalebriety
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 16072009 21:38
(Original post by Tallon)
Yeah sorry about the notation. I don't do this often.
Note that k is positive?
If I drew z+k, wouldn't it be a circle of a radius that could be a mininum of size k (if z is 0) and a max of size 2k (if z = k) centred around (k,0)? arg(z+k) is the angle from the point (k,0) isn't it? The angle all the points of z+k can be from that point? oh ffs i dont know.
That's what I don't understand. How could the argument possibly be between pi/2 and pi/2 if that's the case? 
Tallon
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 16072009 22:00
(Original post by generalebriety)
An equivalent statement to "pi/2 < arg(z+k) < pi/2" is "Re(z+k) > 0".
That's what I don't understand. I would have thought the modulus of z+k would be left of the imaginiary axis centered around k, but it only makes sense if it's right of the axis. I take it that's the answer?
It's also the whole wording that annoys me. I mean wouldn't arg(z+K) be the angle from the point (k,0)? I mean, for example arg(z2) = pi/2 would be a half line upwards from (2,0) wouldn't it? Why is the angle of the argument starting from the origin? Surely it comes from the point (z+K)? 
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 16072009 22:12
(Original post by Tallon)
That's what I don't understand. I would have thought the modulus of z+k would be left of the imaginiary axis centered around k, but it only makes sense if it's right of the axis. I take it that's the answer?
It's also the whole wording that annoys me. I mean wouldn't arg(z+K) be the angle from the point (k,0)? I mean, for example arg(z2) = pi/2 would be a half line upwards from (2,0) wouldn't it? Why is the angle of the argument starting from the origin? Surely it comes from the point (z+K)?
Draw a point in the complex plane representing a complex number z. Then the angle from (k, 0) to this point is arg(z+k). Now, suppose the number z satisfies z <= k; then you are measuring the angle from (k, 0) to some point in the closed disc of radius k about the origin. Can you see that this will always* be between pi/2 and pi/2? (Equivalently, can you see that the point representing z will always be to the right of (k, 0)?)
You can do all of this algebraically, by the way. It's not particularly easy, though. Suppose z <= k; put z = x + iy. Then z^2 = x^2 + y^2 <= k^2, and so z+k^2 = ...
* I don't know what arg(0) is meant to be. I don't think it's defined. The person who set the question obviously overlooked that. 
Tallon
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 16072009 22:41
(Original post by generalebriety)
Draw a point in the complex plane representing a complex number z. Then the angle from (k, 0) to this point is arg(z+k). Now, suppose the number z satisfies z <= k; then you are measuring the angle from (k, 0) to some point in the closed disc of radius k about the origin. Can you see that this will always* be between pi/2 and pi/2? (Equivalently, can you see that the point representing z will always be to the right of (k, 0)?)
You can do all of this algebraically, by the way. It's not particularly easy, though. Suppose z <= k; put z = x + iy. Then z^2 = x^2 + y^2 <= k^2, and so z+k^2 = ...
* I don't know what arg(0) is meant to be. I don't think it's defined. The person who set the question obviously overlooked that.
So for the next part of the question, it's just a simple case of a right angled triangle from (2k,0) to a tangent of the circle and the origin. So the greatest angle it could be is pi/6 and least is pi/6. Greatest distance value of modulus is 3k and least is k?Last edited by Tallon; 16072009 at 22:46. 
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 16072009 23:52
Uh, I'd check what sides of your triangle are what length and what trig function you'd want to use.
Last edited by around; 16072009 at 23:55. 
generalebriety
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 17072009 01:23
(Original post by Tallon)
Yeah I see what you've done there. I didn't quite understand why we were only looking at the points inside the circle of radius k around the origin. But it's because of the first part of the question where it says that modulus of z has to be less than or equal to k right? So only inside that circle is what z can be, yes? So from there on, you can prove what it says with an argand diagram assuming that the modulus of z is less than or equal to k, right?Last edited by generalebriety; 17072009 at 01:27. 
Tallon
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 17072009 15:06
(Original post by generalebriety)
You're confusing yourself. z and k are both just complex numbers. k is real, so we know where k is. The statement "z <= k" means "z lies in the closed disc of radius k about the origin", so we know where z is. The expression "arg(z+k)" means "the angle going from k to z" (or, by symmetry, "the angle going from z to k"). Do you disagree with any of that?
I see, thanks.
so z has to be less than or equal to k, so it can any complex number just so long as it's less than or equal to a distance of k around the origin.
z+k means the distance of z from k? so really it's either a max of the diameter of the circle which is 2k or 0 as it's just on the edge of the circle, right?
arg(z+K) is the angle from k to z, so yeah it would be between (but not including) pi/2 to pi/2. I think I know what you mean about the question being slightly poor because the exact point of (k,0) wouldn't actually be definied there, right? So really, that would is overlooked.
Is all that ok? Thanks a lot. I appreciate that I can be as thick as a plank sometimes but I really appreciate your patience.
(Original post by around)
Uh, I'd check what sides of your triangle are what length and what trig function you'd want to use.Last edited by Tallon; 17072009 at 15:09. 
generalebriety
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 17072009 16:27
(Original post by Tallon)
z+k means the distance of z from k? so really it's either a max of the diameter of the circle which is 2k or 0 as it's just on the edge of the circle, right? 
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 17072009 16:32
ignore this post by order
Last edited by around; 17072009 at 18:28. 
Tallon
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 17072009 18:05
(Original post by around)
Check this. What angle do you want?
I'm pretty sure I'm right in that it's pi/6<=arg(z+2k)<pi/6 because I have the answers in the back of the fp2 book. It's just sine(theta) = k/2k so theta must be pi/6.
(Original post by generalebriety)
Not sure what you mean by this. If z = 9 + 17i, then z+k is the distance from k to 9 + 17i. This quantity doesn't mean anything intrinsically; if you were told that z+k <= k, then you'd know that the distance from k to z was less than or equal to k, and so z would be anywhere in the closed disc of radius k about (k, 0).
Since the modulus of z is less than or equal to K, z can be any complex number in that cirlce, right?
if z is at the point z1 on the diagram, that is, the number k+0i, it's distance from k is 0. If it's at point z2 in the diagram, it's the entire diameter of the circle away from k, and so the distance is 2k. Is that correct?Last edited by Tallon; 17072009 at 18:16. 
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 17072009 18:07
Ok, sorry, I'm making a tit of myself.
feel free to ignore everything i've said 
Tallon
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 17072009 18:18
(Original post by around)
Ok, sorry, I'm making a tit of myself.
feel free to ignore everything i've said
You're really not. Do you have any idea how embarrassing it is asking questions to people on here who are so much better at this than me they must surely think I'm an idiot? lol. 
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 17072009 18:25
Yeah, I see where I went wrong now.
teaches me to post maths at 11 at night after 3 hours fencing :/ 
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 17072009 18:28
(Original post by Tallon)
I agree. I'll try to show what I meant with the imagine attachment. Sorry about how crude it is, lol
Since the modulus of z is less than or equal to K, z can be any complex number in that cirlce, right?
if z is at the point z1 on the diagram, that is, the number k+0i, it's distance from k is 0. If it's at point z2 in the diagram, it's the entire diameter of the circle away from k, and so the distance is 2k. Is that correct?
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