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I need to prove $|z-1| \le ||z|-1|+|z||\arg z|$ using a purely geometric argument.

I know that $|z-1|$ is the distance from z to (1,0). $|z|-1$ is the distance from $z$ to the edge of the unit circle when $z$ is outside of the unit circle. (Similarly, $1-|z|$ is the distance from $z$ to the edge of the unit circle when $z$ is inside of the unit circle.)

I am, however, having a bit of difficulty interpreting what $|z||\arg z|$ is geometrically.

I know that the two vectors $|z-1|$ and $|z|-1$ (or $1-|z|$) seem to form 2 sides of a triangle. I think that $|z||\arg z|$ must somehow be related to that last side of the triangle.

Is $|z||\arg z|$ related to an arc? If so, how is that related to the triangle listed above?

Thanks for all your help in advance!

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2 Answers

up vote 3 down vote accepted

Consider the triangle with vertices $A=1$, $B=|z|$ and $C=z$ on the Argand plane. Then $CA\le AB+BC$. Yet the side length $BC$ is shorter than the length of the circular arc joining $B$ and $C$ on the circle centered at the origin with radius $|z|$. Hence the result.

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Note that $|z||\arg(z)|$ is the length of the arc of the circle centered at $0$ with radius $|z|$ between $z$ and $|z|$ and that $||z|-1|$ is the distance from $|z|$ to $1$. Hence their sum is at least the distance $|z-1|$ between $z$ and $1$ (and the equality holds if and only if $z$ is real nonnegative).

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