What is the maximum value of f(z) for any complex number z?

[juantheron]

If \(\displaystyle z\) is any complex number, Then Maximum value of \(\displaystyle f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|\).

 

[anemone]

If \(\displaystyle z\) is any complex number, Then Maximum value of \(\displaystyle f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|\).

My solution:

Spoiler

If we let $z=x+yi$, we see that

$|z-i|=|x+yi-i|=\sqrt{x^2+(y-1)^2}$

$|z-3-4i|=|x+yi-3-4i|=\sqrt{(x-3)^2+(y-4)^2}$

$|z|=|x+yi|=\sqrt{x^2+y^2}$

$|z-1|=|x+yi-1|=\sqrt{(x-1)^2+y^2}$View attachment 3312

Diagram above shows that we have two triangles, one with the sides lengths of $1,\,|z|,\,|z-i|$ and the other with lengths of $2\sqrt{5},\,|z-1|,\,|z-3-4i|$.

Applying the triangle inequality on both triangles gives

$1+|z|\ge \sqrt|z-i|$ or $1\ge \sqrt|z-i|-|z|$---(1)

and

$2\sqrt{5}+|z-1|\ge |z-3-4i|$ or $2\sqrt{5}\ge |z-3-4i|-|z-1|$ ---(2)

Adding both inequalities (1) and (2) we get

$1+2\sqrt{5}\ge \sqrt|z-i|+|z-3-4i|-|z|-|z-1|$

Hence, the maximum of $f(z)=1+2\sqrt{5}$.

 

[juantheron]

Thanks anemone for Nice Solution.

My solution is same as Yours.(Triangle Inequality.)