Proof of triangle inequality for complex numbers in Baby Rudin, 2nd ed.

Question

In the second edition of Rudin's Principles of Mathematical Analysis he states,

"If $x$ and $y$ are complex numbers, then $|x+y| \le |x|+|y|$."

In the proof of this statement, he starts with,

Put $$\lambda = \frac{|x+y|}{x+y}.$$

Then, multiplying by $x+y$ and using Theorem 1.49b ($|xy|=|x||y|$), we see that $|\lambda|=1$.

I assume that he concludes $|\lambda|=1$ as follows:

If we multiply by $x+y$ we have $(x+y)\lambda=|x+y|$. If I take absolute values on both sides, then by theorem 1.49b, $$|(x+y)\lambda|=|x+y||\lambda|,$$ and on the RHS, $$|(|x+y|)|=|x+y|.$$ Hence, $$|x+y||\lambda|=|x+y|,$$ and so $|\lambda|=1$.

My questions are:

1. When we multiply by $x+y$, we have that $(x+y)\lambda$ (a complex number) equals $|x+y|$ (a scalar). How is this so? I'm just a bit confused by what I see here. In my logic above I took the absolute value of both sides of an equality where, if I'm not mistaken, the LHS is complex and the RHS is strictly real.
2. Is it valid to take the absolute value of $|x+y|$ and conclude that it equals $|x+y|$? I assume so but want to confirm my logic.

Answer 1

When we multiply by x+y, we have that (x+y)λ (a complex number) equals |x+y| (a scalar). How is this so? I'm just a bit confused by what I see here. In my logic above I took the absolute value of both sides of an equality where, if I'm not mistaken, the LHS is complex and the RHS is strictly real.

$\mathbb R \subset \mathbb C$.

So LHS is complex that could be real. And RHS is complex that must be real (and non-negative).

That is not an issue. $\lambda$ was defined in a particular way and as a result we can conclude that $(x+y)\lambda = |x+y|\in \mathbb R^+ $. There was no reason to believe that couldn't happen so we shouldn't be surprised that it did.

We can't expect it of two arbitrary complex numbers but as $\lambda$ was defined by $x+y$ there are certainly not arbitrary.

In other words.... a strictly real number is complex, and a complex number could be strictly real if it was defined to be strictly real.

Is it valid to take the absolute value of |x+y| and conclude that it equals |x+y|? I assume so but want to confirm my logic.

Yes.

$|x+y| = c$ and $c\ge 0$. $(c^2)^{\frac 12}$ is a distinct positive real number $b$ so that $b^2 = c^2$. As $c^2 = c^2$ we can conclude that $(c^2)^{\frac 12} = c$ for all $c\in \mathbb R; c \ge 0$.

Also $c = c+0i = c-0i$ so $c = \overline c$.

So $|c| = (c\overline c)^{\frac 12} = (c^2)^{\frac 12}= c$ for all $c \ge 0$ and .... $|x+y| \ge 0$>

....

or in other words.

Let $x\in \mathbb R$.

If $x \ge 0$ then $|x|= (x\overline x)^{\frac 12} = =[(x + 0i)(x-0i)]^{\frac 12}=(x^2)^{\frac 12}$. As $x \ge 0$ and $x^2 = x^2, $(x^2)^{\frac 12} = x$.

So $|x| = x$ if $x$ is real and $x \ge 0$.

If $x < 0$ then $|x|= (x\overline x)^{\frac 12} = =[(x + 0i)(x-0i)]^{\frac 12}=(x^2)^{\frac 12}$. As $-x > 0$ and $(-x)^2 = x^2, $(x^2)^{\frac 12} = -x$

So $|x| = - x$ if $x$ is real and $x < 0$.

So for real numbers the new-fangled complex absolute value is the same old real absolute value we know and love.

As $|x+y|$ is real and $|x+y| \ge 0$ we have $||x+y|| = |x+y|$.

Answer 2

You can prove this as follows: $$ \begin{align*} |x + y| \leq |x| + |y| &\iff |x + y|^2 \leq (|x| + |y|)^2 \\ &\iff (x + y)\overline{(x + y)} \leq |x|^2 + 2|x||y| + |y|^2 \\ \text{Note: }(x + y)\overline{(x + y)} &= (x + y)(\overline{x} + \overline{y}) \\ &= x \cdot \overline{x} + \overline{x} \cdot y + x \cdot \overline{y} + y \cdot \overline{y} \\ &= |x|^2 + (\overline{x} \cdot y) + \overline{(\overline{x} \cdot y)} + |y|^2 \\ &= |x|^2 + 2Re(\overline{x} \cdot y) + |y|^2 \\ &\leq |x|^2 + 2 |\overline{x}||y| + |y|^2 \\ &= |x|^2 + 2 |x||y| + |y|^2 \end{align*} $$ Which completes the proof.