<s>[INST] 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?
 [/INST] Continue with $|\sqrt 2 z-3+2i|=\frac{1}{\sqrt 2}|z|$ to get,
$$(\sqrt 2 z-3+2i) (\sqrt 2 \bar z-3-2i) =\frac{1}{2}|z|^2$$
or,
$$3|z|^2 -2\sqrt2 (3-2i) \bar z -2\sqrt2 (3+2i) z +26=0$$
which can be written in the form of a circle, i.e.
$$\bigg| z - \frac{2\sqrt2}3(3-2i) \bigg|^2 = \frac{26}9$$
Thus, the locus is a circle with the center at $\ \frac{2\sqrt2}3(3-2i)$ and the radius $\frac{\sqrt{26}}3$.
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