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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.

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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.

enter image description here

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  • $\begingroup$ Can you draw a diagram for a better understanding ? Thanks in advance $\endgroup$ – Mathxx Oct 4 '16 at 17:56
  • $\begingroup$ 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. $\endgroup$ – Brian Tung Oct 4 '16 at 17:59
  • $\begingroup$ @Mathxx: Edited to include picture. $\endgroup$ – Brian Tung Oct 4 '16 at 18:12
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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.)

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