Find the complex numbers $z=x+yi$

Question

$$|\frac{z-4}{z-8}| =1$$ and $$|\frac{z-12}{z-8i}| =\frac{5}{3}$$

This is what I thought to do: $$(z-4)^2/(z-8)^2=1$$ $$-z+2=-2z+8$$ $$-x-yi+2=-2(x+yi)+8$$ $$x=6, y=0.$$ The answer in my book is $6+8i$ or $6+17i$. I don't know how to solve these two equationstogether.

Answer 1

You do not have $|z|^2=z^2$ for complex $z$. Instead $|z|^2=x^2+y^2$ where $z=x+yi$, with $x$ and $y$ being the real and imaginary parts of $z$.

Thus $|z-4|=|z-8|$ implies

$(x-4)^2+y^2=(x-8)^2+y^2$

$x^2-8x+16+y^2=x^2-16x+64+y^2$

The quadratic terms cancel leaving just a linear equation for $x$ which you then solve. Thus $x=6$.

Now put this into the second equation

$|z-12|=(5/3)|z-8i|$

$(x-12)^2+y^2=(25/9)(x^2+(y-8)^2)$

Put $x=6$ into that and solve for $y$ getting the quoted answers from the book.

Answer 2

HINT:Suppose $z=x+iy$.

Now the equations you have becomes $| \frac{x-4+iy}{x-8+iy}|=1$ and $|\frac{x-12+iy}{x+i(y-8)}|$.

Now if you solve this data {rationalise} then you will get that $|(x^2-12x+32+y^2)-i(4y)|=1$ and $|(x^2-1x+y^2-8y)+i(12y+8x)|=5/3$.

Can you take it from here to solve for x and y??

Answer 3

Solve a more general problem, when $\exists\space\text{z}\in\mathbb{C}$ and $\exists\space\text{n}\in\mathbb{R}$:

$$\left|\frac{\text{z}-\text{n}}{\text{z}-2\text{n}}\right|=1\space\Longleftrightarrow\space\left|\text{z}-\text{n}\right|=\left|\text{z}-2\text{n}\right|\space\Longleftrightarrow\space$$ $$\sqrt{\left(\Re\left[\text{z}\right]-\text{n}\right)^2+\Im^2\left[\text{z}\right]}=\sqrt{\left(\Re\left[\text{z}\right]-2\text{n}\right)^2+\Im^2\left[\text{z}\right]}$$

Which leads to:

$$\left(\Re\left[\text{z}\right]-\text{n}\right)^2=\left(\Re\left[\text{z}\right]-2\text{n}\right)^2\space\Longleftrightarrow\space\Re\left[\text{z}\right]=\frac{3\text{n}}{2}$$