SOLUTION: Prove that |z1+z2| < or = |z1| + |z2|. Use |z|^2 = conjugate (z) times z, Re(z) = (z + conjugate (z)) / 2, and Re(z) < or = |z|. This inequality is known as the Triangle Inequality

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Prove that |z1+z2| < or = |z1| + |z2|. Use |z|^2 = conjugate (z) times z, Re(z) = (z + conjugate (z)) / 2, and Re(z) < or = |z|. This inequality is known as the Triangle Inequality      Log On


   

Question 1181369: Prove that |z1+z2| < or = |z1| + |z2|. Use |z|^2 = conjugate (z) times z, Re(z) = (z + conjugate (z)) / 2, and Re(z) < or = |z|. This inequality is known as the Triangle Inequality.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let z%5Bc%5D denote the conjugate of the complex number z.

Then

=

= abs%28z%5B1%5D%29%5E2+%2B+2%2ARe%28z%5B1c%5D%2Az%5B2%5D%29+%2B+abs%28z%5B2%5D%29%5E2, from the fact Re(z) = (z + conjugate (z)) / 2.

, from the fact Re(z) < or = |z|.

===>
Since both abs%28z%5B1%5D%2Bz%5B2%5D%29 and abs%28z%5B1%5D%29%2Babs%28z%5B2%5D%29 are non-negative, the last inequality implies that

abs%28z%5B1%5D%2Bz%5B2%5D%29+%3C=+abs%28z%5B1%5D%29%2Babs%28z%5B2%5D%29, and the Triangle Inequality is proved.