If the point z' is located in the semicircle |z'|<1 and…

If the point $z'$ is located in the semicircle $|z'|<1$ and $|\arg(z')|<\dfrac{\pi}{2}$, show that the point $z=z'^2$ is located in the entire circle $|z|<1$, except for the negative part of the real axis.

Step 1 · Describe the domain

The domain is the open right half-disk:

$$ D = \left\{z' : |z'|<1,\ -\frac{\pi}{2}<\arg(z')<\frac{\pi}{2}\right\} $$

Write $z' = re^{i\theta}$ with $0

Step 2 · Compute $z=z'^2$ in polar form

$$ z = z'^2 = (re^{i\theta})^2 = r^2e^{i2\theta} \qquad\Longrightarrow\qquad |z|=r^2, \quad \arg(z)=2\theta $$

Step 3 · Track the modulus

Since $0

Step 4 · Track the argument

Since $-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}$, doubling gives:

$$ -\pi < 2\theta < \pi \qquad\Longrightarrow\qquad -\pi < \arg(z) < \pi $$

Crucially, both inequalities are strict: $\arg(z)=\pi$ would require $\theta=\dfrac\pi2$, and $\arg(z)=-\pi$ would require $\theta=-\dfrac\pi2$ — but $\theta$ never reaches either endpoint. So $\arg(z)$ takes every value in $(-\pi,\pi)$ but never exactly $\pm\pi$.

Step 5 · Identify the excluded set

The rays $\arg(z)=\pi$ and $\arg(z)=-\pi$ are the same ray: the negative real axis. Since $\arg(z)$ never equals $\pm\pi$, the image never lands on the negative real axis. Combined with Step 3:

$$ \{z=z'^2 : z'\in D\} \subseteq \{z : |z|<1\} \setminus (-1,0] $$

Step 6 · Show every such point is actually reached

Take any $z=\rho e^{i\phi}$ with $0<\rho<1$ and $-\pi<\phi<\pi$ (i.e. any point in the unit disk off the negative real axis). Define:

$$ z' = \sqrt\rho\, e^{i\phi/2} $$

Then $|z'|=\sqrt\rho<1$ ✓, and $\arg(z') = \dfrac{\phi}{2}$ satisfies $-\dfrac{\pi}{2}<\dfrac{\phi}{2}<\dfrac{\pi}{2}$ (since $-\pi<\phi<\pi$) ✓, so $z'\in D$, and by construction $z'^2=z$. So every point of the target set is achieved — the containment in Step 5 is in fact equality.

Step 7 · Check with examples

$z'=\frac1{\sqrt2}e^{i\pi/4}$: $z=z'^2=\frac12e^{i\pi/2}=\frac{i}{2}$ — inside the disk, off the negative real axis. ✓

$z'=0.9$ (on positive real axis): $z=0.81$, still on the positive real axis, closer to $0$.

Attempting $z=-0.5$: solving $z'^2=-0.5=0.5e^{i\pi}$ gives $z'=\sqrt{0.5}\,e^{i\pi/2}=\frac{i}{\sqrt2}$, whose argument $\frac\pi2$ is not strictly inside $\left(-\frac\pi2,\frac\pi2\right)$ — so no $z'\in D$ maps to $-0.5$, confirming the negative real axis is genuinely excluded.