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Answer to Question #5357 in Complex Analysis for richard

Question #5357
if z1 and z2 are the complex numbers show that ||z1|-|z2||<=|z1+z2|<=|z1|+|z2|
Expert's answer
Let's show that | |z1|-|z2| | <= |z1+z2|.
This can be shown very easily on the Argand diagram.& Suppose |z1| > |z2|, but the argument can be shown just as easily if they are the other way around.
Draw in the vector z1. Now from the point z1 draw in the vector z2, (z1 is nose to tail with z2).
Then z1+z2 is the vector joining the origin to the end of the second vector z2.
The distance from the origin to z1 by the direct route is ||z1|-|z2|| + |z2|, whereas going from the origin to z1 via the point (z1+z2) the total distance is |z1+z2| + |z2| and this will be two sides of a triangle and so greater than the direct route.
It follows, comparing these, that
||z1|-|z2|| + |z2| < |z1+z2| + |z2|,
so ||z1|-|z2|| < |z1+z2|.
Equality arises if z1 and z2 are in a straight line with the origin and on opposite sides of the origin.

Let's show that |z1+z2|<|z1|+|z2|.
This is proved with the 'triangle inequality' - two sides of a triangle are always greater than the third side.
In this case we have a polygon which is formed by adding the two vectors nose to tail.
The lefthand side is the magnitude of the straight line joining the origin to the end of the second vector, thereby closing the triangle.
The righthand side is the sum of all sides of the triangle, (except the closing line) and is clearly greater than the single closing line to the end of the second vector.

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