If a>0 and k>1 , show that |a+z| > (a+|z|)/(k) when |arg(z)|…
If $a>0$ and $k>1$, show that $$ |a+z| > \frac{a+|z|}{k} $$ when $\left|\arg(z)\right| < 2\arccos\dfrac{1}{k}$.
Explanation of Method
Write $z = re^{i\theta}$ with $r=|z|$. For fixed $r$, the quantity $|a+z|^2 = a^2+2ar\cos\theta+r^2$ is a strictly decreasing function of $|\theta|$ on $[0,\pi]$ (since $\cos\theta$ decreases there). So the smallest possible value of $|a+z|$ over the allowed range $|\theta|<2\arccos\frac1k$ is approached at the endpoint $|\theta|=2\arccos\frac1k$. It suffices to show the inequality holds (with equality allowed) exactly at that endpoint — then every interior $\theta$, having strictly larger $\cos\theta$, gives a strictly larger $|a+z|$, which is the strict inequality we want.
Step 1 · Set up in polar form
Let $z = re^{i\theta}$, $r=|z|\ge 0$. Then:
$$ |a+z|^2 = |a+r\cos\theta + ir\sin\theta|^2 = (a+r\cos\theta)^2 + (r\sin\theta)^2 = a^2 + 2ar\cos\theta + r^2 $$
The inequality to prove, $|a+z| > \dfrac{a+r}{k}$, is equivalent (both sides positive) to:
$$ a^2 + 2ar\cos\theta + r^2 > \frac{(a+r)^2}{k^2} $$
Step 2 · Reduce to the boundary angle
Let $\varphi = \arccos\dfrac1k$, so $\cos\varphi = \dfrac1k$ and the allowed range is $|\theta| < 2\varphi$. Since $2\varphi \in (0,\pi)$ and $\cos\theta$ is strictly decreasing on $[0,\pi]$, for every $\theta$ with $|\theta|<2\varphi$ we have $\cos\theta > \cos(2\varphi)$, hence:
$$ a^2+2ar\cos\theta+r^2 \;>\; a^2+2ar\cos(2\varphi)+r^2 $$
So it is enough to show the right-hand side is $\ge \dfrac{(a+r)^2}{k^2}$; the strict inequality above then finishes the proof.
Step 3 · Evaluate at $\theta = 2\varphi$
Using the double-angle formula with $\cos\varphi = \dfrac1k$:
$$ \cos(2\varphi) = 2\cos^2\varphi - 1 = \frac{2}{k^2}-1 = \frac{2-k^2}{k^2} $$
so:
$$ a^2+r^2+2ar\cos(2\varphi) = a^2+r^2+\frac{2ar(2-k^2)}{k^2} $$
Multiplying out and regrouping over the common denominator $k^2$:
$$ = \frac{k^2a^2+k^2r^2+4ar-2ark^2}{k^2} = \frac{k^2(a^2+r^2-2ar)+4ar}{k^2} = \frac{k^2(a-r)^2+4ar}{k^2} $$
$$ \boxed{a^2+r^2+2ar\cos(2\varphi) = (a-r)^2 + \frac{4ar}{k^2}} $$
Step 4 · Compare to the target value $\dfrac{(a+r)^2}{k^2}$
We need $(a-r)^2 + \dfrac{4ar}{k^2} \ge \dfrac{(a+r)^2}{k^2}$. Multiply through by $k^2>0$:
$$ k^2(a-r)^2 + 4ar \ge (a+r)^2 $$
Since $(a+r)^2 - 4ar = (a-r)^2$, this is equivalent to:
$$ k^2(a-r)^2 \ge (a-r)^2 \quad\Longleftrightarrow\quad (k^2-1)(a-r)^2 \ge 0 $$
which is true for every real $a,r$, since $k>1 \Rightarrow k^2-1>0$ and $(a-r)^2 \ge 0$. So:
$$ a^2+r^2+2ar\cos(2\varphi) = (a-r)^2+\frac{4ar}{k^2} \;\ge\; \frac{(a+r)^2}{k^2} $$
with equality exactly when $a=r$.
Step 5 · Combine
Chaining Steps 2 and 4, for every $\theta$ with $|\theta| < 2\varphi = 2\arccos\dfrac1k$:
$$ a^2+2ar\cos\theta+r^2 \;>\; a^2+2ar\cos(2\varphi)+r^2 \;\ge\; \frac{(a+r)^2}{k^2} $$
so $|a+z|^2 > \dfrac{(a+r)^2}{k^2}$, and since both sides are non-negative, taking square roots gives:
$$ \boxed{|a+z| > \frac{a+|z|}{k}} $$
The first inequality in the chain is strict for every such $\theta$ (even when $a=r$, where the second inequality becomes an equality), so the overall inequality is always strict on the stated range. $\blacksquare$
The allowed directions for $z$ (shaded cone, half-angle $2\varphi=2\arccos\frac1k$ on each side of the positive real axis, shown for $k=1.5$). $|a+z|$ is largest along $\arg z=0$ and smallest — but still $>\frac{a+|z|}{k}$ — as $\arg z$ approaches the cone's edge.
Result
For $a>0$, $k>1$, and $z$ with $|\arg z| < 2\arccos\dfrac1k$:
The proof reduces the problem to the boundary angle $\theta = 2\varphi$ (where $\varphi=\arccos\frac1k$), evaluates $|a+z|^2$ there in closed form as $(a-r)^2+\dfrac{4ar}{k^2}$, and shows this always meets the target bound via the identity $(k^2-1)(a-r)^2 \ge 0$ — with strict inequality inherited from $\cos\theta > \cos(2\varphi)$ for every $\theta$ strictly inside the cone.