# Least value of $|z-w|$

On an argand diagram, sketch the locus representing complex numbers $z$ satisfying $|z+i|=1$ and the locus representing complex numbers $w$ satisfying arg$(w-2)=\frac{3}{4}\pi$ find the least value of $|z-w|$ for points on these loci.

I have sketched the loci. And can anyone teach me how to find the least value with a diagram? Thanks

• Can you describe the equations of the loci? – thanasissdr Apr 2 '16 at 8:06

$\qquad\qquad\qquad$ See the above diagram. Here, note that $\triangle{ABC},\triangle{AOD}$ and $\triangle{DEF}$ are triangles with $45^\circ,45^\circ,90^\circ$ and with $|AC|=|AO|=|OD|=1,|ED|=|EO|-|OD|=2-1=1$.

Thus, \begin{align}\text{the least value of |z-w|}&=|CF|\\&=|CD|+|DF|\\&=|AD|-|AC|+\frac{1}{\sqrt 2}|DE|\\&=\color{red}{\frac{3}{2}\sqrt 2-1}\end{align}

• @Mathxx : The least value of $|z−w|$ is the shortest distance between a point on the circle represented by $z$ and a point on the line represented by $w$. So, we consider a line perpendicular to the line and passing through the center of the circle $A$. I hope that this makes it easier to understand the answer. – mathlove Apr 3 '16 at 7:15

The shortest distance between two curves is along a common normal. Since one of the curves here is a line, it is easy to give an equation of its normal at each point. Then we just have to see which one is a normal to the other curve (a circle) as well. The minimum distance should be along this line.

Here is a diagram showing what's happening. The solid line from the circle and the line is the one with the least distance. It's also a common normal. The dotted lines show what happens to the distance if you perturb this line slightly.