3
$\begingroup$

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 $|a|\geq2.$

$\endgroup$
3
$\begingroup$

$|k|=|-a-k-b+k+a+b-k|\le |a+k| + |b-k| + |a+b-k|=1$ 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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.