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A question related to this does exists before. Main point: When triangle inequality in complex numbers is applicable? Is there a condition? Or this is applicable for all complex numbers? If so, why we have the following ambiguity?

Here's the question,

If a complex number $z$ lies in the interior or on the boundary of a circle of radius $3$ units and centre $(– 4, 0)$, find the greatest and least values of $|z +1|$ .

So, if we visualize a diagram, we get the biggest complex arrow $z$ = $-7$

Therefore, the maximum value is $|-7+1| = 6$

But, if we go with the triangle inequality,

$$|z+1|\leq |z| + 1\leq 7+1 = 8$$

So, why the maximum value of $|z+1|$ is $8$ now.

Where I went wrong with the triangle inequality?

Robert Z
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Ariana
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  • Triangle inequality does not provide always the maximum of the given expression, but just an upper bound of this... – dmtri Oct 06 '19 at 08:01

1 Answers1

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There is no further condition to apply triangle inequality with complex number. In your case, for $z=-7$, it holds, because $6\leq 8$. Nothing went wrong!

Note that if you take $z=7$ (which is outside your disc) then $$8=|7+1|=|z+1|\leq |z|+1 =7+1=8.$$ In your disc, equality holds for example when $z=-7$ and $w=-1$, then $$8=|(-7)+(-1)|=|z+w|\leq |z|+|w|=|-z|+|-w|=8.$$ Please see also Equality of triangle inequality in complex numbers

Robert Z
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