# Finding minimum value of the equation given.

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.

• $|1-z| = |-(z-1)|=|z-1|$ – krirkrirk Nov 16 '18 at 12:42
• @krirkrirk Thanks! It didn't click at the moment. Got the answer! – Kaustuv Sawarn Nov 16 '18 at 12:47

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.

$$|1+z|+|1-z|=c$$ 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, $$c\ge 2$$.

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 , $$z=?$$