If a,b are complex numbers, k its an integer, $k \neq 0$ and $|a+k| + |b-k| + |a+b-k|=1$ then proof that $a,b$ are real numbers

I've tried

$a+k=x$ and $b-k=y$

Then I used absolute value inequalities $|x|+|y|+|x+y-k|\geq |x|-|y|+|x+y|-|-k| \geq -k \implies k\geq-1$

On the other hand $|x|+|y|+|x+y-k|=|x|+|y|+|k-(x+y)|\geq|x|+|y|+|k|-|x|-|y|=k \implies 1\geq k$ Hence, I found that $k\in \{-1;1\}$ and i have some ideas how to prove that $2\geq|a|$ and if I could prove that $|a|\geq2$ then $|a|=2$. That will be more easy but I dont know how to find that $|a|\geq2.$


$|k|=|-a-k-b+k+a+b-k|\le |a+k| + |b-k| + |a+b-k|=1$ so $|k| \le 1$. Since $k \ne 0$ integer, it follows $|k| \ge 1$, hence $|k|=1$

$1=|a+k| + |b-k| + |-a-b+k| \ge |a+k|+|a| \ge |a+k-a| =|k|=1$

so we have equality in all the inequalities and in particular $|a+k|+|a|=|k|$ which means immediately $a$ real and of opposite sign to $k$

(either geometrically, or just square $|a|^2+|k|^2+2k\Re a =|a+k|^2= (|a|-|k|)^2=|a|^2+|k|^2-2|a||k|$, so $|a|=-sgn(k)\Re(a)$

But now $1=|a+k| + |b-k| + |-a-b+k| \ge |a+b|+|-a-b+k| \ge |k|=1$, so we have equalities everywhere, in particular $|a+b-k|+|a+b|=|k|$, so as above $a+b$ real and of the same sign to $k$, hence $b$ real since $a$ real!

  • $\begingroup$ Thank you for helping! $\endgroup$ – mathematiciangrade8 May 14 at 17:47
  • $\begingroup$ you are welcome! $\endgroup$ – Conrad May 14 at 17:58

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