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Question - Find the minimum value of $|1 + z| + |1-z|$.

I'm trying to solve the question by thinking of them as points in the Argand plane. The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane. But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.

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  • $\begingroup$ $|1-z| = |-(z-1)|=|z-1|$ $\endgroup$ – krirkrirk Nov 16 '18 at 12:42
  • $\begingroup$ @krirkrirk Thanks! It didn't click at the moment. Got the answer! $\endgroup$ – Kaustuv Sawarn Nov 16 '18 at 12:47
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The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.

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$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.

BEcause the axis must be longer than the distance between focus, $c\ge 2$.

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A bit of geometry.

In the complex plane:

$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.

$\triangle ABC$ has sides of lengths:

$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.

The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:

$|z+1|+|z-1| > 2.$

$2$ is a lower bound. Is there a minimum?

If yes , $z=?$

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