Find the locus of the complex number from the given data If $|\sqrt 2 z-3+2i|=|z||\sin (\pi/4+\arg z_1) + \cos (3\pi/4-\arg z_1)|$, where $z_1=1+\frac{1}{\sqrt 3} i$, then the locus of $z$ is ...
$$\arg~z_1 =\frac{\pi}{6}$$
So RHS of the equation is $\;|z|\dfrac{1}{\sqrt 2}$.
Then 
$\;|\sqrt 2 z-3+2i|=|z|\dfrac{1}{\sqrt 2}$.
How should I proceed?

$Arg[1+i/\sqrt{3}]= \pi/6$
$$|z\sqrt{2}-3-2i|=|z||\sin(\pi/4+\pi/6)+cos(3\pi/4-\pi/6)|$$
$$\implies |z\sqrt{2}-3-2i|=|z|\sqrt{\frac{1}{2}} \implies
\left|\frac{2z-(3+2i)\sqrt{2}}{z}\right|=1$$
The locus of $z$ is a circle. Let $z=x+iy$,
we have $$(2x-3\sqrt{2})^2+(2y-2\sqrt{2})^2=(x^2+y^2)$$
tHe simplified eq. of circle is
$$(x-2 \sqrt{2})^2+(y-4\sqrt{2}/3)^2=10$$
