More complex numbers~~ (1 Viewer)

[pLuvia]

Given that z is a variable complex number such that |z-2-2i| = 2, show on an Argand diagram, the locus of the point P which represents z.

Hence, or otherwise, find
(i) the greatest value of |z|
(ii) z in the form z=x+iy when arg(z-2-2i) = 2pi/3

I've done the first bit, but can't get (i) or (ii)

 

[YBK]

|z-(-2+2i)| = 2

then sketch the thing.. the centre will be the point (-2,1).... that wil be the centre of a circle with radius 2. Find the equation of circle by using the distance formula i think... it will be (x+2)^2 + (y-1)^2 = 4

then extend a line from the origin through the point (-2,1), that will touch the circle at two ends, the close and far away one. after that is done, all you are left to do is find the far away point to get the max value. find modulus of -2+2i and add that to 2.

 

[Riviet]

Hey man, i'm here to help you of course ^^

i) Greatest value occurs when z is furthest away from the origin
.: max |z|=2+2+sqrt(2²+2²)
=4+sqrt8

 

[*yooneek*]

*claps*!!!!!!!!!!!
ur so smart!!!

awesuuuuummmmmmm!!!

 

[YBK]

*claps*!!!!!!!!!!!
ur so smart!!!

awesuuuuummmmmmm!!!

just a love for maths...

and not as smart as you

 

[pLuvia]

Isn't centre (2,2)
|z-2-2i| = 2
|(x-2)+(y-2)i| = 2
root (x-2)²+ (y-2)² = 2
|(x-2)²+ (y-2)²| = 2

Centre (2,2) radius 2
Well that's what it says in the answers

 

[YBK]

Isn't centre (2,2)
|z-2-2i| = 2
|(x-2)+(y-2)i| = 2
root (x-2)²+ (y-2)² = 2
|(x-2)²+ (y-2)²| = 2

Centre (2,2) radius 2
Well that's what it says in the answers

:s why..........

 

[mitsui]

yea centre (2,2)
as it is actualli
|z-(2+2i)|=2

 

[YBK]

Isn't centre (2,2)
|z-2-2i| = 2
|(x-2)+(y-2)i| = 2
root (x-2)²+ (y-2)² = 2
|(x-2)²+ (y-2)²| = 2

Centre (2,2) radius 2
Well that's what it says in the answers

lol, wait

yeah that's right

|z-2-2i|
= |z - (2 + 2i) |

yup yup it's hard doing maths on teh computer

 

[YBK]

Hey man, i'm here to help you of course ^^

i) Greatest value occurs when z is furthest away from the origin
.: max |z|=2+2+sqrt(2²+2²)
=4+sqrt8

Isn't it 2+sqrt8 ?

 

[pLuvia]

Sorry to tell this guys lol, both of you are wrong hehe

Riviet: Thanks for your help it triggered by senses I remember how to do it again hehe

Well if you want the correct answer

Origin - O
Centre - A
Furthest from origin - P

OA = 2root2 ( Root(4+4))
AP = 2 (radius)
.: OA + AP = OP
OP = |z| = 2+2root2
= 2(1+root 2)

thanks for your help lol

 

[pLuvia]

Isn't it 2+sqrt8 ?

Yep .

 

[YBK]

Yep .

hehe, that means i got part (i) right at least

 

[SeDaTeD]

The maximum also occurs when the slope of the tangent of the circle (x-2)^2 + (y-2)^2 = 4 is equal to the slop of the tangent of the circle x^2 + y^2 = r^2, for some r. This also gives the point of minimum distance. You can see why this is so because if at a point the two slopes were different, the circles would intersect so there are some points on the circle in question outside of the circle centred at the origin.

for the circle at the origin:
2x + 2y. dy/dx = 0
dy/dy = -x/y

for the circle in question:
2(x-2) + 2(y-2).dy/dx = 0
dy/dx = -(x-2)/(y-2)

therefore -x/y = -(x-2)/(y-2)
x/y = (x-2)/(y-2)

but this condition says that the lines joining thesolution point to the origin and (2,2) must have the same slope. This occurs only the line passes through both the origin and (2,2). The the solution points are the intersection of this line through (2,2) , being y=x, and the circle. thus (x-2)^2 + (x-2)^2 = 4
(x-2)^2 = 2
x-2 = +/- sqrt2
x = 2 +/- sqrt2 => y = 2 +/- sqrt 2
This gives the solutions for the nearest (2-sqrt2,2-sqrt2) and furthest (2+sqrt2,2+sqrt2)points from the origin.

 

[pLuvia]

aa, so smart lol

 

[YBK]

The maximum also occurs when the slope of the tangent of the circle (x-2)^2 + (y-2)^2 = 4 is equal to the slop of the tangent of the circle x^2 + y^2 = r^2, for some r. This also gives the point of minimum distance. You can see why this is so because if at a point the two slopes were different, the circles would intersect so there are some points on the circle in question outside of the circle centred at the origin.

for the circle at the origin:
2x + 2y. dy/dx = 0
dy/dy = -x/y

for the circle in question:
2(x-2) + 2(y-2).dy/dx = 0
dy/dx = -(x-2)/(y-2)

therefore -x/y = -(x-2)/(y-2)
x/y = (x-2)/(y-2)

but this condition says that the lines joining thesolution point to the origin and (2,2) must have the same slope. This occurs only the line passes through both the origin and (2,2). The the solution points are the intersection of this line through (2,2) , being y=x, and the circle. thus (x-2)^2 + (x-2)^2 = 4
(x-2)^2 = 2
x-2 = +/- sqrt2
x = 2 +/- sqrt2 => y = 2 +/- sqrt 2
This gives the solutions for the nearest (2-sqrt2,2-sqrt2) and furthest (2+sqrt2,2+sqrt2)points from the origin.

*is confused*

for the circle at the origin:
2x + 2y. dy/dx = 0 (that part doesnt make sense to me... maybe cause we haven't completly finished the dirivatives topic yet i guess)


isn't it easier to draw the circle and then see ?

 

[SeDaTeD]

That's not in 4unit I don't think, but it's not hard to understand.

oh and YBK, i used implicit differentiation on the equation x^2 + y^2 = r^2. You may not have done that yet. Basically you differentiate the equation in its implicity form, rather than rewriting it as an equation of y = f(x)

Oh and yes, I think it can also be done purely geometrically, without having to use any differentiation. Draw the circles such that they just touch at the furthest ppint, and you can construct lines joining their centres and use some geometric and symmetry areguements to say hy it's the furthest point.

 

[YBK]

That's not in 4unit I don't think, but it's not hard to understand.

yeah, the concept makes total sense - pretty cool!

 

[YBK]

how can you do the second part of that question...? i don't think i get that..

 

[YBK]

oh and YBK, i used implicit differentiation on the equation x^2 + y^2 = r^2. You may not have done that yet. Basically you differentiate the equation in its implicity form, rather than rewriting it as an equation of y = f(x)

Oh and yes, I think it can also be done purely geometrically, without having to use any differentiation. Draw the circles such that they just touch at the furthest ppint, and you can construct lines joining their centres and use some geometric and symmetry areguements to say hy it's the furthest point.

ahhh.. okay I see.. we haven't done that yet.

Yeah, I solved it geometrically - it's how we were taught how to solve it.