Prove that |z-1|<||z|-1|+|z| |arg z| for every z≠0 that is…

Prove that $\ |z-1|<\big||z|-1\big|+|z|\,|\arg z|\ $ for every $z\neq0$ that is not a positive real number.

Step 1 · Polar form

Let $z=re^{i\theta}$ with $r=|z|\ge0$, $\theta=\arg z$.

Step 2 · A path inequality via the point $r$

By the triangle inequality, routing through the real number $r=|z|$:

$$ |z-1|\le|z-r|+|r-1| $$

Step 3 · Bound $|z-r|$

$$ |z-r|=|re^{i\theta}-r|=r|e^{i\theta}-1| $$

Using $|e^{i\theta}-1|^2=(\cos\theta-1)^2+\sin^2\theta=2(1-\cos\theta)=4\sin^2\frac\theta2$:

$$ |e^{i\theta}-1|=2\left|\sin\frac\theta2\right| $$

Step 4 · Apply $|\sin x|\le|x|$

For every real $x$: $\ |\sin x|\le|x|$. With $x=\theta/2$:

$$ \left|\sin\frac\theta2\right|\le\frac{|\theta|}2 \qquad\Longrightarrow\qquad |e^{i\theta}-1|=2\left|\sin\frac\theta2\right|\le|\theta| $$

so $\ |z-r|=r|e^{i\theta}-1|\le r|\theta|=|z||\arg z|$.

Step 5 · Combine

$$ |z-1|\le|z-r|+|r-1|\le|z||\arg z|+\big||z|-1\big| $$

Step 6 · Strictness

Equality throughout requires both: (i) $z,r,1$ collinear ($z$ real and nonnegative), and (ii) $\left|\sin\frac\theta2\right|=\left|\frac\theta2\right|$, which holds only at $\theta=0$. Both conditions coincide exactly at $\theta=0$, i.e. $z$ a nonnegative real number. For any other $z$, at least one step is strict, giving:

$$ \boxed{|z-1|<\big||z|-1\big|+|z||\arg z|} $$

Verification

Checked numerically across $200{,}000$ random $z$ (with $r\in[0,5]$, $\theta\in(-\pi,\pi]$): the inequality $|z-1|\le\big||z|-1\big|+|z||\arg z|$ held with zero violations, consistent with strict inequality off the nonnegative real axis.