least value of |w − z| On an Argand diagram, sketch the loci of points representing complex numbers $w$ and $z$ such that $|w − 1 − 2i|= 1$ and $\arg(z-1)=\dfrac{3}{4}\pi$.
Find the least value of $|w-z|$ for points on these loci. 
My attempt, I've already drawn the loci of w. But I don't know how to draw the arg one.  Can anyone explain it to me how to draw and how to find the least value. Thanks a lot.
 A: Basic approach. The locus of $\arg (z-1) = \frac{3\pi}{4}$ is that set of points $z$ for which a ray drawn from the point represented by $1$ on the complex plane, through $z$, makes an angle of $\frac{3\pi}{4}$ with the real axis.
That should also tell you which point on the locus of $w$ is the closest to the locus of $z$: Draw a line from $1+2i$, perpendicular to the ray representing the locus of $z$.  Where that line intersects the loci of $w$ and $z$ are the desired points.

A: We want, as is evident from the picture,  the point on $\operatorname{arg}(z-1)=\dfrac {3\pi}4$ ( $y=-x+1$) which is closest to the center,$(1,2)$, of the circle.
It's easy to see that $\operatorname{arg} z=\dfrac{3\pi}4 $ is the line $y=-x$.  Now just replace $x$ by $x-1$ to get $y=-x+1$ for $\operatorname{arg}(z-1)=\dfrac{3\pi}4$.
So, the line through $(1,2)$ and perpendicular to $y=-x+1$ will intersect $y=-x+1$ at the closest point.  The equation of that line is $y=x+1$.  We get $(0,1)$ as the point on $y=-x+1$ closest to the circle. 
Plugging $y=x+1$ into the equation of the circle and solving,  we get that $(1-\dfrac {\sqrt2}2,2-\dfrac{\sqrt2}2)$ is the point on the circle closest to the line  $y=-x+1$.  (The problem didn't ask for this info though.)
At any rate,  we now have two ways to compute the minimum distance. One is to compute the distance between $(0,1)$ and $(1,2)$;  then subtract the radius.  Get $\sqrt2-1$.
Or $\operatorname{min}\mid w-z\mid=\mid  i-((1-\dfrac {\sqrt2}2)+(2-\dfrac {\sqrt2}2)i)\mid=\sqrt{3-2\sqrt2}=\sqrt2-1$.
(Of course,  that these two will come out the same is perfectly trivial.)
