# Find all complex numbers satisfying $z\cdot\bar{z}=41$, for which $|z-9|+|z-9i|$ has the minimum value

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!

• Geometrically speaking the least value would occur when an ellipse with $(9,0)$ and $(0,9)$ is tangent to the given circle. Symmetry suggests that one such point $\left(\sqrt{\frac{41}{2}}, \sqrt{\frac{41}{2}}\right)$. And since it is tangent at this point to the circle, this is the unique point. Jul 5, 2018 at 11:38

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.

• Sorry for being stupid, but how do I express the condition of ellipse being tangent to the circle in terms of algebra? Jul 5, 2018 at 9:56

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.

• Can you please elaborate algebraic steps? Jul 5, 2018 at 10:38

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.