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

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The maximum value of the function f(z) defined as f(z) = |z - i| + |z - 3 - 4i| - |z| - |z - 1| is explored in the discussion. The solution relies on the Triangle Inequality to derive the maximum value effectively. Participants express agreement on the approach and solutions presented. The discussion emphasizes the importance of understanding complex number properties in solving such problems. Overall, the focus remains on maximizing the function within the constraints of complex analysis.
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If $$z$$ is any complex number, Then Maximum value of $$f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|$$.
 
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jacks said:
If $$z$$ is any complex number, Then Maximum value of $$f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|$$.

My solution:

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 $f(z)=1+2\sqrt{5}$.
 

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Thanks anemone for Nice Solution.

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