Min. value of sum of two complex no.

[man111]

If [tex]z = x+iy[/tex] is a complex no. then find Min. value of [tex]|2z-1|+|3z-2|[/tex]

[martin.nikolov]

It looks easy.

[tex]|2z-1|+|3z-2|=|(2x-1)+iy|+|(3x-2)+iy|=\sqrt{(2x-1)^2+y^2}+\sqrt{(3x-2)^2+y^2}[/tex]

which is greater than or equal to

[tex]\sqrt{(2x-1)^2}+\sqrt{(3x-2)^2}=|2x-1|+|3x-2|[/tex]

with equality when y=0. Now find the min of this function.

[drago]

[tex]|2z-1|+|3z-2|=6(|z-\frac{1}{2}|+|z-\frac{2}{3}|) \ge \ 6(|\frac{2}{3}-\frac{1}{2}|)=1.[/tex]
The equality takes place when [tex]z \in [\frac{1}{2}, \frac{2}{3}].[/tex]

[martin.nikolov]

[tex]|2z-1|+|3z-2|=6(|z-\frac{1}{2}|+|z-\frac{2}{3}|) \ge \ 6(|\frac{2}{3}-\frac{1}{2}|)=1.[/tex]
The equality takes place when [tex]z \in [\frac{1}{2}, \frac{2}{3}].[/tex]

!!!

[mkmarinov]

Doesn't seem so right to me...
Let [tex]z=\frac{1}{2}+0i[/tex]
[tex]|2z-1|+|3z-2|=|2.\frac{1}{2}-1|+|3. \frac{1}{2}-2|=|0|+|-\frac{1}{2}|=\frac{1}{2}[/tex]

[man111]

Here is My Solution.

Put [tex]z = x+iy[/tex], and Rewrite The expression as [tex]2|z-\frac{1}{2}|+3|z-\frac{2}{3}|[/tex]We Get

[tex]K = 2\sqrt{(x-\frac{1}{2})^2+y^2}+3\sqrt{(x-\frac{2}{3})^2+y^2}[/tex]

and Let [tex]a = \sqrt{(x-\frac{1}{2})^2+y^2}[/tex] and [tex]b=\sqrt{(x-\frac{2}{3})^2+y^2}[/tex]

Then [tex]K=2a+3b...............................(1)[/tex](Here We have to Minimize that sum.)

Now [tex]a(\frac{1}{2},0)[/tex] and [tex]b(\frac{2}{3},0)[/tex] are the Distance from point [tex]P(x,y)[/tex] in Cartesian plane.

So for Min. Distance Points [tex]P,a[/tex]and [tex]b[/tex]lie on Straight line i.e on Real axis(X-axis).

So [tex]a+b=\frac{1}{6}......................(2)[/tex] So [tex]a=\frac{1}{6}-b[/tex]

Put in equ...(1), We Get

[tex]K=2(\frac{1}{6}-b)+3b=\frac{1}{3}+b=\frac{1}{3}+\sqrt{(x-\frac{2}{3})^2+y^2}\geq \frac{1}{3}[/tex]

So [tex]K_{Min}=\frac{1}{3}[/tex]

[drago]

Oops, what a slip! Sorry.
But, still, the same idea could work.

[tex]|2z-1|+|3z-2|=2(|z-\frac{1}{2}|+ |z-\frac{2}{3}| ) + |z-\frac{2}{3}| \ge 2|\frac{2}{3}-\frac{1}{2}| + |z-\frac{2}{3}| \ge \frac{1}{3}.[/tex]
Because:
[tex](1) |z-\frac{1}{2}|+ |z-\frac{2}{3}| \ge |\frac{2}{3}-\frac{1}{2}|.[/tex]
(triangle's inequality which becomes equality when [tex]z \in [\frac{1}{2}, \frac{2}{3}][/tex])
[tex](2) |z-\frac{2}{3}| \ge 0[/tex]
The equality takes place when both (1) and (2) do, i.e. when z=2/3