As $\arg (a\times b)=\arg a\arg b$ the equation becomes:

$\arg z-\arg (z-(1-2i))=\pi /3$

The rest of the answer gets somewhat geometrical, so look at the diagram.

Draw a triangle $OAB$, with $O$ at the origin and $B$ at $(1,-2)$. Let $A$ represent $z$. Draw lines $OX$, $Y\text{'}BY$ parallel to the $x$-axis as shown.

Then $\arg z=XOA$ (where $XOA$ represents the angle between $XO$ and $OA$) and $\arg (z-(1-2i))=YBA$. Therefore the equations can be written

$XOA-YBA=\pi /3$

As $BOX=Y\text{'}BO$ we can write

$(BOA-Y\text{'}BO)-YBA=\pi /3$

$BOA-(Y\text{'}BO+YBA)=\pi /3$

By angles on a straight line ( $Y\text{'}OY$) this becomes:

$BOA-(\pi -ABO)=\pi /3$

and by angles in a triangle:

$OAB=\pi /3$

(Note I have been careful to ensure all angles are described in an anticlockwise sense)

So the angle $OAB$ is constant, and as all angles subtended by the same arc of a circle are equal $A$ lies on an arc of a circle connecting $O$ to $B$, such that the size of the angle $\left|OAB\right|$ is $\pi /3$. There are two such arcs, one below $O$ and $B$ and one above.

One arc will correspond to the $\pi /3$ and the other $-\pi /3$, so how do we find out which is which? Well here's a rather crude way:

For $A$ to be above $OB$ clearly $YBA\ge XOA$ (measured as arguments i.e. in an anticlockwise sense). Therefore $\arg z-\arg (z-(1-2i))<0$ so does not equal $\pi /3$.

So $A$ must lie below $OB$. So we have determined which arc the locus of $z$ lies on.

I hope this helps - please get back to me if it makes no sense.

Yours,