wildareal's questions thread

[brightsky]

Don't entirely understand your question but I'll assume k = r...

Split the 11z^2 up:

z^4 - 2z^3 + [/b]2z^2 + 9z^2[/b] -18z + 18 = 0

Factorise:

z^2(z^2 - 2z + 1) + 9(z^2 - 2z + 1) = 0

(z^2 + 9)(z^2 - 2z + 1) = 0

Use null factor law, you have the following complex solutions: -3i, 3i, 1+i and 1 - i. Hence k = r = -3 or 3.

[kamil9876]

Another way:

plugging in gives:





Since is real we can seperate this into imaginary and real parts. Looking at the imaginary part you see:


so this means r can only be either: , test which of these also satisfy the real part (remember, both real and imaginary parts must be 0), we see 0 doesn't work, but 3 and -3 do, hence they are the ones.

[wildareal]

If a and b are a complex numbers, Im(a)=2, Re(b)=-1 and a+b=-ab, find a and b.

Thanks.

[jasoN-]

Let
    , where x and y are real numbers






Equating real coefficients:



Equating imaginary coefficients:

as



hence
and

[wildareal]

Describe z: mod(z+1)=mod(z-i)

Thanks.

[brightsky]

z = x + yi

|x+1 + yi| = |x + (y-1)i|

sqrt((x+1)^2 + y^2) = sqrt(x^2 + (y - 1)^2)

x^2 + 2x + 1 + y^2 = x^2 + y^2 - 2y + 1

2x = -2y

x = -y

Hence Re(z) = -Im(z)

[taiga]

Wildareal can you please make one thread with all your questions. I don't think it's necessary for you to make a new thread each question you have.

And Brightsky, I'm thoroughly impressed that you can do that whilst being in year 9.

[luken93]

And Brightsky, I'm thoroughly impressed that you can do that whilst being in year 9.

Go to the super fun happy maths thread in the spesh board and then you will see what he can do haha

[wildareal]

A complex number z satisfies the inequality mod(z+2-2root(3)i)≤2

a) Find the least value of mod(z)

b) Find the greatest possible value of Arg(z)

[_avO]

someone should merge all these threads into one

[jasoN-]

a circle at point with radius 2

The minimum value of mod(z) is the smallest distance from the origin to the circle

(look at picture)

To find x, simply use Pythagoras' theorem.



So minimum value = Distance from origin to center - Radius
                         = 4 - 2
                         = 2



For your next question I'm not too sure, but I'll give it a go.
(full speculating here lol.)

Max value of Arg(z) is the angle from the positive x-axis to the point on the circle producing the largest angle. (ie. a tangent)
Looking at pic2, and knowing that a tangent to a circle produces a right angle with the radius from the center we can find the angle.
From previous question we have x = 4
the angle between the red and green line (lol) is:








I'm not entirely sure on either these answers, need confirmation.

[evaever]

Describe z: mod(z+1)=mod(z-i)

Thanks.

It is a set of complex numbers such that they are 'equidistant' from z=-1 and z=i in the complex plane. They lie on the perpendicular bisector of a line segment joining z=-1 and z=i.

[wildareal]

Question is attached Thanks.

[brightsky]

a) XY = 1/2 DA + a - 1/2 CB [1], XY = b - 1/2 DA + 1/2 CB [2]
[1] + [2]
2XY = a + b
XY = (a+b)/2
b) a and b are parallel, so that means b = k_1. a, where k_1 is scalar. For XY to be parallel to AB. (a + b)/2 = 1/2 a + 1/2 b = k_2.a, where k_2 is some scalar.
Sub b = k_1. a into the above.
1/2 a + k_1 a = (1/2 + k_1) a.
Since 1/2 + k_1 is scalar, then XY is parallel to a = AB.

[evaever]

a)  OX=(OA+OD)/2, OY=(OB+OC)/2, XY=OY-OX=(OB-OA+OC-OD)/2=(AB+DC)/2=(a+b)/2

b)  a//b, so b=ka, XY=(a+ka)/2=((1+k)/2)a=((1+k)/2)AB, so XY//AB