Answer to Question #215974 in Complex Analysis for James

Question

Find the smallest value of P=|z-2|^2 + |z+1-i|^2 + | z-2 -5i|
And z (z = x +yi : x,y are real numbers) is a complex numbers satisfies the condition 2|(x+yi)-1-2i| = |3i + 1 - 2(x-yi)|

Answer 1

Let us develop the expression of $P=|z-2|^2+|z+1-i|^2+|z-2-5i|$ in terms of $x$ and $y$ :

$P=(x-2)^2+y^2+(x+1)^2+(y-1)^2+\sqrt{(x-2)^2+(y-5)^2}$

And the condition is $4(x-1)^2+4(y-2)^2=(1-2x)^2+(3+2y)^2$.

The condition can be simplified (after developing all the squares) as

$-4x-28y+10=0$

or $2x+14y=5$

Now as this condition admits a reformulation in the form $x=7y-2.5$, we can substitue this into the expression of $P$ :

$P=(7y-4.5)^2+y^2+(7y-1.5)^2+(y-1)^2+\sqrt{(7y-4.5)^2+(y-5)^2}$

Which we can rewrite as

$P=100y^2-86y+23.5+\sqrt{50y^2-73y+45.25}$

Now by calculating $P'=200y-86+\frac{100y-73}{2\sqrt{50y^2-73y+45.25}}$

By finding the root and submitting it into the expression of $P$ yields us that the minimum of $P$ in the given conditions is $P(x_{min},y_{min})\approx 9.7934$