Complex number equation to cartesian equation

Question

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...

Answer 1

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 $5$.