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According to wolfram alpha, this inequality holds:

$$|1-z| \leq |z|+1,$$

however, I haven't been able to figure out why. The reverse inequality (or at least the way I've applied it) gives me a lower bound on $|1-z|$ instead of an upper bound. I've tried using the regular triangular inequality and adding and subtracting like $|1-z+z-1| \leq |1-z| + |z-1|$ but I'm still stuck.

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    $\begingroup$ $$|1-z| \le |1|+|-z|$$ $\endgroup$ – Crostul Jul 4 '18 at 7:07
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By the triangle inequality $$|z|+1=|z|+|-1|\geq|z-1|=|1-z|.$$

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$$|1-z|\leqslant|1|+|-z|=1+|z|.$$

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Hint: You are right to think of the triangle inequality, but only need to recall that $$|z|=|-z|,$$ by definition.

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A bit of geometry:

Draw vectors from (0,0) in the complex plane:

$\vec 1$, and $\vec z$.

Join the endpoint of $\vec z$ with the endpoint of $\vec 1$ resulting in vector $\overrightarrow {(1-z)}.$

Consider this triangle:

Side lengths : $|1-z|$, $1$, and $|z|.$

The sum of the lengths of $2$ sides of a triangle is greater than the $3$rd side.

Hence: $|1-z| \lt |z| +1.$

For $|z| =0$ : $|1-z|=|z| +1$, equality.

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