# 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 $arg(z-1)=\frac{3}{4}\pi$.

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.

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.

• Can you draw a diagram for a better understanding ? Thanks in advance – Mathxx Oct 4 '16 at 17:56
• Gee whiz, so demanding! :-P All right, I'll sketch it out, but leave out the actual values; those you'll have to determine on your own. – Brian Tung Oct 4 '16 at 17:59
• @Mathxx: Edited to include picture. – Brian Tung Oct 4 '16 at 18:12

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.)