Proof some complex number inequality?

Question

Proof that, for all z, w an element of C (complex numbers);

| z + w | <= | z | + | w |

(<= means less or equal to).

Thanks for the help.

Answer 1

The proof of that inequality is better represented graphically:
http://en.wikipedia.org/wiki/Image:Triangle_inequality.svg

As you can see, the sum of any two sides of a triangle has to be greater than or equal to the third side. I understand that your question is not about triangles and geometry, but understand that the absolute value of real numbers, complex numbers, etc. can "represent," (be attributed to), any two sides of that triangle.

For example, if we let ||x|| = |z|; ||y|| = |w|; and ||x + y|| = |z + w|, the you may be able to see that |z| + |w| >= |z + w|.
If you have trouble picturing the proof, then draw out your own triangles on a sheet of paper, and assign |z|, and |w| to ANY two sides of the triangle. You'll find that the third side, |z + w|, is always less than the sum of the two other sides.

Again, this is applicable to triangles (geometry), real numbers and complex numbers. The reason being that the sides of a triangle can represent the absolute values of real and complex numbers.

Answer 2

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Answer 3

A few preliminary results: First, if z* denotes the conjugate of z, then (z+w)* = z*+w* and z*w* = (zw)*. These can be verified by manual calculation -- let z=x+yi and w=a+bi, then z*+w* = x-yi + a-bi = (x+a) - i(y+b) = (z+w)* and z*w* = (x-yi)(a-bi) = (xa - yb) - i(ya + xb) = (zw)*.

Next, observe that |z|² = x² + y² = (x+iy)(x-iy) = zz*. So per the above, |z|²|w|² = zz*ww* = zwz*w* = zw(zw)* = |zw|², and taking square roots of both sides yields that |z||w| = |zw|.

Finally, note that z+z* = x+iy + x-iy = 2x = 2Re(z), Re(z) = x ≤ |z|, |z*| = √(x²+(-y)²) = √(x²+y²) = |z|, and (z*)* = z.

So now for the actual proof:

|z+w|²
= (z + w)(z + w)*
= (z + w)(z* + w*)
= zz* + wz* + zw* + ww*
= |z|² + wz* + (z*)*w* + |w|²
= |z|² + wz* + (wz*)* + |w|²
= |z|² + 2Re(wz*) + |w|²
≤ |z|² + 2|wz*| + |w|²
= |z|² + 2|w||z*| + |w|²
= |z|² + 2|w||z| + |w|²
= (|z| + |w|)²

So we have that |z+w|² ≤ (|z| +|w|)², and taking square roots of both sides yields the triangle inequality. Q.E.D.