My first attempt was to express $z$ as $x+iy$ and minimize the expression $\sqrt{(x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}$ where $x^2+y^2=41$.
That said, it seems to me that using the geometric interpretation could be easier. As far as I understand, I need to find points on the circle for which the sum of distances to the points $(9,0)$ and $(0,9)$ is lowest. This interpretation, however, doesn't help with regard to calculations.
Is there some simple trick or idea I'm missing?
Thank you!
On
Hint: Show that $$\sqrt{(8x-9)^2+y^2}+\sqrt{x^2+(y-9)^2}\geq 9\sqrt{2}$$ and the equal sign holds if $$x=4,y=5$$ ok we will prove the inequality above: squaring all we get
$$2\sqrt{(8x-9)^2+y^2}\sqrt{(x^2+(y-9)^2}\geq 162-(8x-9)^2-(y-9)^2-x^2-y^2$$, now we use that $x^2+y^2=41$: we get
$$2\sqrt{(8x-9)^2+41-x^2}\sqrt{41-y^2+(y-9)^2}\geq 121-(8x-9)^2-(y-9)^2$$
squaring again and factorizing we get
$$- \left( 64\,{x}^{2}+{y}^{2}-144\,x-18\,y+41 \right) \left( 64\,{x}^{ 2}+{y}^{2}-144\,x-18\,y+42 \right) \geq 0$$ which is true.
On
Apply the triangle inequality to the triangle with vertices $z, 9, 9i$:
$|z-9|+|z-9i|=|-(z-9)|+|z-9i| \ge |9-9i|$
with equality occurring only when $z$ lies on the line segment between the fixed points $9, 9i$. If this line segment intersects the circle, then all such points of intersection forcibly minimize the left side of the above inequality.
You should get whole numbers for the minimizing real and imaginary parts.
The locus of points with sum of distances $a$ from $(9,0)$ and $(0,9)$ is an ellipse. If we have $a=9\sqrt{2},$ we get a degenerate line segment between the 2 points, but as $a$ increases, the ellipse expands and then becomes tangent to the circle. Thus, you want to find the value of $a$ so that the ellipse with foci at $(9,0)$ and $(0,9)$ is tangent to the circle $x^2+y^2=41.$ Upon finding $a,$ the point of tangency is the desired $z.$
Having completed the interpretation, I leave the calculation to you.