# Complex numbers problem with absolute value property

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!