TITLE: Complex number equation to cartesian equation
QUESTION [1 upvotes]: I need to find the cartesian equation of the loci of z given the below equation:
$\frac{(z-j)\cdot(z-j)^*}{|3-4j|}=5$
So if $z=x+jy$
Then $(z-j)^*=(x+jy-j)^*$
$=[x+j(y-1)]^*$
$=x-j(y-1)$
$=x-jy+y$
Substituting back into original equation
$\frac{(x+jy-j)\cdot(x-jy+j)}{|3-4j|}=5$ 
$\frac{(x+j(y-1))\cdot (x-j(y-1))}{|3-4j|}=5$
$\sqrt{x^2+(y-1)^2}\cdot \sqrt{x^2-(y-1)^2}=5|3-4j|$
This is where I'm getting a little lost...

REPLY [1 votes]: For any complex number $w = a + bj$,
$$|w| = \sqrt{a^2 + b^2}$$
$$ww^* = |w|^2 = a^2 + b^2$$
hence if $z=x + yj$, then 
\begin{align*}
z-j &= x +yj - j = x + (y-1)j\\[4pt]
\implies\; (z-j)(z-j)^*&=|z-j|^2\\[4pt]
&= |x + (y-1)j|^2\\[4pt]
&= x^2 + (y-1)^2\\[4pt]
\end{align*}
Also, $|3-4j| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5,\,$ hence
\begin{align*}
&\frac{(z-j)(z-j)^*}{|3-4j|}=5\\[6pt]
\implies\; &\frac{x^2 + (y-1)^2}{5}=5\\[6pt]
\implies\; &x^2 + (y-1)^2=25\\[6pt]
\end{align*}
which is the equation of a circle in the $xy$-plane, centered at $(0,1)$, with radius $5$.
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