The minimum value of $|z-1+2i| + |4i-3-z|$ is

Question

The minimum value of $$|z-1+2i| + |4i-3-z|$$ is?

The only method of moving further that comes to my mind is assuming $$z=x+iy$$.

Answer 1

Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.

Answer 2

The minimum is the distance between $1-2i$ and $3-4i$ which is 4$\sqrt {(3-1)^2+(-4+2)^2} = 2\sqrt 2 $$

This is when $z$ is on the segment joining the two points and $z$ is between them.

Answer 3

You may proceed as follows:

You have

• $|z-a| + |z-b| \stackrel{!}{\rightarrow} \mbox{Min}$ with
• $a= 1-2i$ and $b = -3+4i$ The triangle inequality gives immediately $$|z-a| + |z-b| \geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2\sqrt{13}$$

Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.

Answer 4

A bit of geometry in the complex plane:

1)$d:=$

$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.

$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.

1) $A,B,C$ are not collinear.

In $\triangle ABC:$

$d= |AC|+|BC| >|AB|.

(Strict triangle inequality ).

2) $A,B,C$ are collinear.

a) $z$ is within the line segment $AB$,

then $d=|AB|$((why?).

b) $z$ is outside the line segment $|AB|$,

then $d>|AB|$(why?).