If $z$ is a complex number satisfying the equation $|z+i|+|z−i|=8 $ then maximum value of $|z|$ is? If $z$ is a complex number satisfying the equation $|z+i|+|z−i|=8 $ then maximum value of $|z|$ is ?
I took $z$ as a point p on the graph and drew lines connecting it to $i$ and $-i$. I assumed $z=x+iy$. Therefore x and y should be maximum. If x and y are maximum, the triangle by i,-i and z has maximum area i.e. height is maximum. Taking i and -i as the base, max height come out when z is on the x axis at distance √15 from origin. But the answer is 4. Please solve it.
 A: What do you understand by $|z+i|+|z-i|=8$ ?
It's basically the sum of distances of a point from the points $(0,i)$ and the point $(0,-i)$ on the imaginary axis. The sum is a constant. What does that mean?
A: $|z+i|+|z−i|=8$ is an ellipse with foci at $z=i$ and $z=-i$. The points with maximal modulus are at the end of the major axis, that is $z= \pm 4i$.
Or with pure arithmetic:
$$
 2|z| = |(z-i) + (z+i)| \le |z+i|+|z−i|=8 \\
\implies |z| \le 4
$$
with equality for $z= \pm 4i$.
A: Apply Minkowski Inequality:
$8 = |z-i| + |z+i| = \sqrt{x^2+(y-1)^2} + \sqrt{x^2+ (y+1)^2} \ge \sqrt{(x+x)^2 + ((y-1)+(y+1))^2} = \sqrt{(2x)^2 + (2y)^2} = 2\sqrt{x^2+y^2} = 2|z| \implies |z| \le 4 \implies |z|_{\text{max}} = 4$ , when $x = 0, y = \pm 4 \implies z = \pm 4i.$
A: Note that $$|Z_1+Z_2| \le |Z_1|+ |Z_2|$$
So $$8=|z+i|+|z-i| \ge |z+i+z-i| \implies 8 \ge |2z| \implies |z| \le 4.$$
Also geometrically $|z+i|+|z-i|=2a$ represents an ellispes whosesemi major axis is $a=4$ this also gives the maximum possible value of $|z|$ (maximum distance between origin and any point on ellipse).
OP should realize that complex numbers do not admit max, min or inequality. It $|z|$ that is real and has max.
