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# P6 Complex numbers watch

1. This is Ex. 2C, Question 12 of the Heinemann book (page 43) if anyone's interested. I think I'll just type out the whole question, but please note I only need help on the last part (unless one of the other parts is wrong, of course).

(i) Show that in an Argand diagram the equation,

arg(z-2) - arg(z-2i) = 3pi/4, represents an arc of a circle.

(ii) Show that |z-4|/|z-1| is constant on this circle.

(iii) Find the values of z corresponding to the points in which this circle is cut by the curve given by,

|z-1| + |z-4| = 5

================================ =================

(i) Circle with equation x²+y²=4
I think the arc required is the one such that x+y > 2 (see bolded part of curve in attachment).

(ii) The expression wittles down to 4, i.e. a constant.

(iii) Now, I think this is in the form of an equation of an ellipse, with centre (5/2 , 0), and foci at 1 and 4 on the x-axis. The ellipse passes through (0,0) and (5,0) and is at a maximum at (5/2 , ½√91). Is that correct?
I now wanted to get the equation of the ellipse in its cartesian form (so I could solve), but realised I didn't know how to do it with the centre shifted.
Any help is appreciated!
Attached Images

2. (ii) should be 2. That is:
|z-4|/|z-1| = 2
|z-4| = 2|z-1|

By substituting this in (iii), you should get:
3|z-1| = 5
|z-1| = 5/3

Which is a circle with center (1, 0) and radius 5/3, i.e. it's equation is:
(x-1)² + y² = 25/9
x² - 2x + 1 + y² = 25/9
4 - 2x + 1 = 25/9 --- (since x²+y²=4)
x = 10/9

Use this to solve for y.

Oh, and I think your sketch is correct.
3. Fantastic! I failed to see the connection between the second and third parts.

Thanks very much.
4. (Original post by mockel)
This is Ex. 2C, Question 12 of the Heinemann book (page 43) if anyone's interested. I think I'll just type out the whole question, but please note I only need help on the last part (unless one of the other parts is wrong, of course).

(i) Show that in an Argand diagram the equation,

arg(z-2) - arg(z-2i) = 3pi/4, represents an arc of a circle.

================================ =================

(i) Circle with equation x²+y²=4
I think the arc required is the one such that x+y > 2 (see bolded part of curve in attachment).
From an algebraic point of view you can do it as follows.

Remember tan(A-B) = (tanA-tanB)/(1+tanA tanB)

So

tanLHS = [y/(x-2) - (y-2)/x]/(1+y(y-2)/{x(x-2)}) = -1 = tanRHS

yx - (x-2)(y-2) +x(x-2) + y(y-2) = 0

yx - xy + 2y + 2x - 4 + x^2 + y^2 -2x -2y = 0

x^2 + y^2 = 4

which answers the question.

But if you want to think about it geometrically it's better to think about this equation

arg(z-2) - arg(z-2i) = 3pi/4

as a circular arc connecting 2 and 2i which makes a right angle at its centre (from circle theorems) and so the centre is O by inspection
5. Ah, I actually did it the algebraic way, exactly as you've done it, because I was majorly confused as to what it would look like.
Thanks for the tip for the geometrical way :)
6. arg(z-2) - arg(z-2i) = 3pi/4,

anyone know what that would look like? been puzzled for a few hours now! and exams on tuesday!!! - i thought it would be an arc on the opposite side - i.e. angle subtended being pi/4 in the 3rd quadrant.

any clues? - pic would be very appreciated

phil
7. The pic is attached in mockel's post...
8. (Original post by dvs)
The pic is attached in mockel's post...
why is it that - anyone got any logical explanation? - i cant see how that can enclose an angle of 3pi/4

i get it now - how dumb of me, lol - i was confused about what happens when an angle is > pi rad - anyone know what happens tehn, and will we get a question where the arg of two things like that question below = say 3pi/2? etc?
9. It would be the other arc. (See pic - it's the red arc. Btw, this isn't meant to be the exact same pic as mockel's, but I was too lazy to make a new one. The part in black would now represent arg... = pi/2.)

The reasoning behind this is that we want pi+pi/2, i.e. pi/2 subtended in the 'other' side.
Attached Images

10. (Original post by dvs)
It would be the other arc. (See pic - it's the red arc. Btw, this isn't meant to be the exact same pic as mockel's, but I was too lazy to make a new one. The part in black would now represent arg... = pi/2.)

The reasoning behind this is that we want pi+pi/2, i.e. pi/2 subtended in the 'other' side.
If you sub z =-2 (which is on your arc) into

arg(z-2) - arg(z-2i)

you get

pi - (-3pi/4) = 7pi/4

which is incorrect (even up to multiples of 2pi)

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