Let 0<a<1 . Prove analytically and geometrically that if…

Let $0<a<1$. Prove analytically and geometrically that if $|z|<a$, then $$ |\arg(1+z)| < \arcsin a. $$

Step 1 · Analytical setup

Let $z=x+iy$ with $x^2+y^2<a^2$. Then $1+z=(1+x)+iy$.

Step 2 · Stay in the right half-plane

Since $|z|<a<1$, we have $|x|<1$, so $1+x>0$. Thus $1+z$ lies strictly to the right of the imaginary axis, and $$ \arg(1+z)\in\Bigl(-\frac{\pi}{2},\frac{\pi}{2}\Bigr). $$

Step 3 · Bound the sine of the argument

$$ \sin\bigl(\arg(1+z)\bigr) = \frac{y}{|1+z|} = \frac{y}{\sqrt{(1+x)^2+y^2}}. $$

We claim $$ \left|\frac{y}{\sqrt{(1+x)^2+y^2}}\right| < a. $$ Squaring both sides (which are non-negative) is equivalent to $$ y^2 < a^2\bigl[(1+x)^2+y^2\bigr] \quad\Longleftrightarrow\quad y^2(1-a^2) < a^2(1+x)^2. $$

Step 4 · Use the disk constraint

From $x^2+y^2<a^2$ we have $y^2<a^2-x^2$. Since $1-a^2>0$, it is enough to prove $$ (a^2-x^2)(1-a^2) \le a^2(1+x)^2. $$ Expanding and simplifying reduces this to $$ 0 \le a^4+2a^2x+x^2 = (a^2+x)^2, $$ which is always true. Moreover $a^2+x>0$ under $|x|<a<1$, so the inequality is strict.

Step 5 · Analytical conclusion

Hence $|\sin(\arg(1+z))|<a$. Because $\arg(1+z)$ lies in $\bigl(-\frac{\pi}{2},\frac{\pi}{2}\bigr)$ and $\sin$ is strictly increasing there, $$ |\arg(1+z)| < \arcsin a. $$

Step 6 · Translate the region

The condition $|z|<a$ is the open disk of radius $a$ about $0$. Adding $1$ shifts it to the open disk $$ |w-1|<a $$ centered at the point $1$ on the real axis.

Step 7 · Tangent lines from the origin

Draw the two lines from the origin $O$ tangent to the circle $|w-1|=a$. In the right triangle $OCT$ formed by $O$, the center $C=1$, and a tangency point $T$: $$ CT \perp OT, \qquad |OC|=1, \qquad |CT|=a, $$ so $$ \sin(\angle TOC) = \frac{a}{1} = a \quad\Longrightarrow\quad \angle TOC = \arcsin a. $$

Step 8 · Containment in the tangent cone

Those two tangents bound a sector of half-angle $\arcsin a$ around the positive real axis. The open disk $|w-1|<a$ lies strictly inside that sector, so every $w=1+z$ in the disk satisfies $$ |\arg(w)| = |\arg(1+z)| < \arcsin a, $$ matching the analytical result.