On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers

Question

Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrically, when the strict inequality would hold. Whenever, however, would we have equality? Would it be necessary that $\lvert x \rvert = \rvert y \rvert$, or is there some other necessary condition?

Thanks.

Answer 1

Let $x=a+bi, y=c+di,$ where $a,b,c,d \in \mathbb{R}.$ Then $$|x+y|=|(a+c)+(b+d)i|=\sqrt{(a+c)^2+(b+d)^2},$$ and $$|x|+|y|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}.$$ Thus, the inequality is equivalent to $$\sqrt{(a+c)^2+(b+d)^2} \leq \sqrt{a^2+b^2}+\sqrt{c^2+d^2},$$by squaring both sides, namely,$$ac+bd \leq \sqrt{(a^2+b^2)(c^2+d^2)}.$$ But in fact, by Cauchy's inequality, we have $$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2.$$ Hence,$$\sqrt{(a^2+b^2)(c^2+d^2)} \geq \sqrt{(ac+bd)^2}=|ac+bd|\geq ac+bd.$$

The equality holds if and only if $ad=bc$ and $ac+bd \geq 0.$

Answer 2

\begin{align*} &|x + y| = |x| + |y| \\ \implies \, &|x + y|^2 = (|x| + |y|)^2 \\ \implies \, &|x + y|^2 = |x|^2 + |y|^2 + 2|x||y| \\ \implies \, &(x + y)\overline{(x + y)} = x\overline{x} + y\overline{y} + 2|x||y| \\ \implies \, &x\overline{x} + y\overline{y} + x\overline{y} + y\overline{x} = x\overline{x} + y\overline{y} + 2|x||y| \\ \implies \, &x\overline{y} + \overline{x\overline{y}} = 2|x||y| \\ \implies \, &2\Re(x\overline{y}) = 2|x\overline{y}| \\ \end{align*} So, we have a complex number $z$ such that $\Re(z) = |z|$. If $z = a + ib$, then $$a = \sqrt{a^2 + b^2} \implies a^2 = a^2 + b^2 \implies b^2 = 0 \implies b = 0,$$ hence $z$ is real. Moreover, $z$ must be positive. Therefore, $x\overline{y} \in [0, \infty)$. The argument of $x\overline{y}$ is $0$, hence the difference between the arguments of $x$ and $y$ is $0$. That is, $x$ and $y$ must lie on the same ray from the origin, and therefore must be a positive real multiple of each other.

Answer 3

We may take a somewhat general approach which follows the spirit of the comment of anon to the question itself.

First we observe that $\Bbb C$ is possessed of a real inner product $\langle \cdot, \cdot \rangle_{\Bbb C}$, viz. for $z_1, z_2 \in \Bbb C$,

$\langle z_1, z_2 \rangle_{\Bbb C} = \Re(z_1 \bar z_2); \tag 1$

if we write

$z_1 = x_1 + iy_1, \; z_2 = x_2 + iy_2, \tag 2$

then

$z_1 \bar z_2 = (x_1 + i y_1)(x_2 - iy_2) = (x_1 x_2 + y_1 y_2) + i(x_2 y_1 - x_1 y_2), \tag 3$

whence

$\langle z_1, z_2 \rangle_{\Bbb C} = \Re(z_1 \bar z_2) = x_1 x_2 + y_1 y_2 = (x_1, y_1) \cdot (x_2, y_2), \tag 4$

which shows that $\Bbb C$ may be considered as an inner product space over $\Bbb R$, and that the inner product $\langle z_1, z_2 \rangle_{\Bbb C}$ is identical with the usual inner product on the real vector space $\Bbb R^2$; in particular, for $z = x + iy$, the modulus $\vert z \vert_{\Bbb C}$ given by

$\vert z \vert_{\Bbb C}^2 = \langle z, z \rangle_{\Bbb C} \tag 5$

is identical with the norm $\vert \cdot \vert_{\Bbb R}$ corresponding to the $\cdot$-product on $\Bbb R^2$:

$\vert z \vert_{\Bbb C}^2 = \langle z, z \rangle_{\Bbb C} = \Re (z \bar z) = x^2 + y^2 = (x, y) \cdot (x, y) = \vert (x, y) \vert_{\Bbb R}^2; \tag 6$

that $\langle \cdot, \cdot \rangle_{\Bbb C}$ is in fact a bilinear form over $\Bbb R$ is easily verified and the details are left to the reader. We thus see that $\Bbb C$ with $\langle \cdot, \cdot \rangle_{\Bbb C}$ is a real inner product space.

Now let $V$ be a real inner product space with inner product $\langle \cdot, \cdot \rangle_V$ and norm

$\Vert v \Vert_V^2 = \langle v, v \rangle_V \; \text{for} \; v \in V; \tag 7$

if $x, y \in V$ with

$y = \alpha x, \; 0 \le \alpha \in \Bbb R, \tag 8$

then

$\Vert x + y \Vert_V = \Vert x + \alpha x \Vert_V = \Vert (1 + \alpha)x \Vert_V = (1 + \alpha) \Vert x \Vert_V$ $= \Vert x \Vert_V + \alpha \Vert x \Vert_V = \Vert x \Vert_V + \Vert \alpha x \Vert_V = \Vert x \Vert_V + \Vert y \Vert_V; \tag 9$

now suppose

$\Vert x + y \Vert_V = \Vert x \Vert_V + \Vert y \Vert_V; \tag{10}$

then we have

$\Vert x + y \Vert_V^2 = \langle x + y, x + y \rangle_V = \langle x, x \rangle_V + \langle x, y \rangle_V + \langle y, x \rangle_V + \langle y, y \rangle_V$ $ = \Vert x \Vert_V^2 + 2\langle x, y \rangle_V + \Vert y \Vert_V^2; \tag{11}$

also,

$(\Vert x \Vert_V + \Vert y \Vert_V)^2 = \Vert x \Vert_V^2 + 2\Vert x \Vert_V \Vert y \Vert_V + \Vert y \Vert_V^2; \tag{12}$

then by virtue of (10) we find

$\Vert x \Vert_V^2 + 2\langle x, y \rangle_V + \Vert y \Vert_V^2 = \Vert x \Vert_V^2 + 2\Vert x \Vert_V \Vert y \Vert_V + \Vert y \Vert_V^2, \tag{13}$

whence

$\langle x, y \rangle_V = \Vert x \Vert_V \Vert y \Vert_V; \tag{14}$

we now invoke the Cauchy-Schwarz inequality which affirms that

$\vert \langle x, y \rangle_V \vert \le \Vert x \Vert_V \Vert y \Vert_V, \tag{15}$

with equality precisely when $x$ and $y$ are collinear; that is, when $y = \beta x$ for some $\beta \in \Bbb R$; now since each side of (14) is non-negative, we see that it implies

$\vert \langle x, y \rangle_V \vert = \Vert x \Vert_V \Vert y \Vert_V; \tag{16}$

but then

$\beta \Vert x \Vert_V^2 = \beta \langle x, x \rangle_V = \langle x, \beta x \rangle_V = \langle x, y \rangle_V = \Vert x \Vert_V \Vert y \Vert_V = \Vert x \Vert_V \Vert \beta x \Vert_V = \vert \beta \vert \Vert x \Vert_V^2, \tag{17}$

whence, if $x \ne 0$,

$\beta = \vert \beta \vert \ge 0, \tag{18}$

and we see that (10) implies that $y$ is a non-negative multiple of $x$ in any real inner product space $V$.

Returning now to our original problem, the case $V = \Bbb C$, we see that, since (1) defines a real inner product on $\Bbb C$, we have

$\vert z_1 + z_2 \vert_{\Bbb C} = \vert z_1 \vert_{\Bbb C} + \vert z_2 \vert_{\Bbb C} \tag{19}$

precisely when

$z_2 = \alpha z_1 \tag{20}$

for some $0 \le \alpha \in \Bbb R$.