Proof: (13) Assume $|x|≥|y|$. Then we must show $|x|-|y| > |x-y|$ is impossible. Multiply both sides by (positive) $|x|+|y|$ to obtain $|x|^2-|y|^2 > |x-y|(|x|+|y|) ≥ |x-y||x+y| = |x^2-y^2|$ so $|x^2| > |x^2-y^2|+|y^2| ≥ |x^2|$, a contradiction. The case is parallel when $|y|≥|x|$.
(14) Then $z=a+bi$ such that $a^2+b^2=1$. We see $|1-z|=|z-1|$ and now $|z+1|^2+|z-1|^2=(a+1)^2+b^2+(a-1)^2+b^2=2(a^2+b^2)+2=4$.
(15) Let $v_a,v_b∈\mathbb{C}^k$ be the the vectors of the $a_j$ and $b_j$. Let $A = \sum |a_j|^2$,$B = \sum |b_j|^2$, and $C = \sum a_j\overline{b_j}$. We claim $|C|^2=AB$ if and only if $v_b$ is associate to $v_a$ (i.e. $v_b=cv_a$ for some $c∈\mathbb{C}$) or at least one of $v_a$ or $v_b$ is zero. ($⇒$) Assume $|C|^2=AB$ and $v_a,v_b \neq 0$ so $A,B > 0$. As we have seen, $∑|Ba_j-Cb_j|^2 = B(AB-|C|^2)$. When $|C|^2=AB$ we then thus have $a_j=\dfrac{C}{B}b_j$ for all $j$, and thus $v_b = \dfrac{C}{B}v_a$. ($⇐$) When either one of $v_a$ or $v_b$ is zero the equality clearly holds, so assume $v_a,v_b \neq 0$ and $v_b = cv_a$ for some complex $c$. We can thus manipulate$$B=\sum |b_j|^2 = \sum |ca_j|^2 = |c|^2\sum |a_j|^2 = |c|^2A$$and$$\overline{c}A=\overline{c}\sum a_j\overline{a_j} = \sum a_j\overline{b_j} = C$$so for all $j$ we have$$Ba_j = |c|^2Aa_j = \dfrac{|c|^2a_j\overline{a_j}A}{\overline{a_j}} = \dfrac{|ca_j|^2}{\overline{a_j}}A = b_j \dfrac{\overline{b_j}}{\overline{a_j}} A = b_j \overline{c} A = Cb_j$$so that $∑ |Ba_j - Cb_j|^2 = B(|C|^2 - AB) = 0$. Since $v_b \neq 0$ and thus $B > 0$, we have $|C|^2 = AB$.$~\square$
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