# Thread: Complex Numbers - Equation

1. ## Complex Numbers - Equation

Hi guys, I'm stumped on this, and I was hoping someone could help me out - so here goes:

Find all the complex numbers to the equation: 2|z| = |z-3-3i|.

2. Originally Posted by takamura
Hi guys, I'm stumped on this, and I was hoping someone could help me out - so here goes:

Find all the complex numbers to the equation: 2|z| = |z-3-3i|.
The locus is an Apollonian circle (but the geometric significance of the relation may not mean anything to you).

Substitute $\displaystyle z = x + iy$:

$\displaystyle 2 |x + iy| = |(x - 3) + (y - 3)i|$

Calculate the magnitudes in the usual way. Then square both sides and simplify.

3. wrong work .. sorry . Thanks Prove it.

4. Originally Posted by mathaddict
Let $\displaystyle z=a+bi$

$\displaystyle 2|a+bi|=|a+bi-3-3i|$

$\displaystyle 2|a+bi|=|a-3+(b-3)i|$

$\displaystyle 2(a+bi)^2=[a-3+(b-3)i]^2$ $\displaystyle \color{red}|a + bi| \neq (a + bi)^2$ and $\displaystyle \color{red}|(a - 3) + (b - 3)i| \neq [(a - 3) + (b - 3)i]^2$.

$\displaystyle 2(a^2+2abi-b^2)=[(a-3)^2+2(a-3)(b-3)i-(b-3)^2]$

$\displaystyle a^2-b^2+6a-6b+(2ab+6a+6b-18)i=0$

SO $\displaystyle a^2-b^2+6a-6b=0$ ---1

and $\displaystyle 2ab+6a+6b+18=0$ ----2

Now we have to solve the simultaneous equation

From 1

$\displaystyle (a-b)(a+b+6)=0$

$\displaystyle a=b$ , $\displaystyle a=-6-b$

Substitute into 2

$\displaystyle ab+3a+3b-9=0$

so when $\displaystyle a=b$ , $\displaystyle b^2+6b-9=0\Rightarrow b=-3\pm3\sqrt{2}$

and a will be the same .

when $\displaystyle a=-6-b$ ,$\displaystyle (-6-b)b+3(-6-b)+3b-9=0$

$\displaystyle b^2-6b-27=0$

b=9 or b=-3 , so you can find a

so you should get 4 different sets of z .

PLs check through my algebra work .
Should actually be

$\displaystyle 2|a + bi| = |(a - 3) + (b - 3)i|$

$\displaystyle 2\sqrt{a^2 + b^2} = \sqrt{(a - 3)^2 + (b - 3)^2}$

$\displaystyle 4(a^2 + b^2) = (a - 3)^2 + (b - 3)^2$

$\displaystyle 4a^2 + 4b^2 = a^2 - 6a + 9 + b^2 - 6b + 9$

$\displaystyle 3a^2 + 6a + 3b^2 + 6b = 18$

$\displaystyle a^2 + 2a + b^2 + 2b = 6$

$\displaystyle a^2 + 2a + 1 + b^2 + 2b + 1 = 8$

$\displaystyle (a + 1)^2 + (b + 1)^2 = (2\sqrt{2})^2$.

So the solution is the circle of radius $\displaystyle 2\sqrt{2}$ units, centred at $\displaystyle (-1, -1)$.