Find the locus of the complex number from the given data

Question

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?

Answer 1

$Arg[1+i/\sqrt{3}]= \pi/6$ $$|z\sqrt{2}-3-2i|=|z||\sin(\pi/4+\pi/6)+cos(3\pi/4-\pi/6)|$$ $$\implies |z\sqrt{2}-3-2i|=|z|\sqrt{\frac{1}{2}} \implies \left|\frac{2z-(3+2i)\sqrt{2}}{z}\right|=1$$ 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 $$(x-2 \sqrt{2})^2+(y-4\sqrt{2}/3)^2=10$$

Answer 2

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 $\frac{\sqrt{26}}3$.