For which value of the real parameter α is the number z_1 =…

For which value of the real parameter $\alpha$ is the number $z_1 = 1+i\alpha$ closer to the origin than the number $z_2 = 1+\dfrac{i}{\alpha}$?

Step 1 · Compute $|z_1|$

$$ |z_1| = |1+i\alpha| = \sqrt{1+\alpha^2} $$

Step 2 · Compute $|z_2|$

(requiring $\alpha \neq 0$):

$$ |z_2| = \left|1+\frac{i}{\alpha}\right| = \sqrt{1+\frac{1}{\alpha^2}} $$

Step 3 · Set up the inequality

$$ \sqrt{1+\alpha^2} < \sqrt{1+\frac{1}{\alpha^2}} $$

Both sides are positive, so squaring is valid:

$$ 1+\alpha^2 < 1+\frac{1}{\alpha^2} $$

Step 4 · Simplify

Subtracting $1$ and multiplying by $\alpha^2>0$ (which preserves the inequality direction):

$$ \alpha^2 < \frac{1}{\alpha^2} \quad\Longrightarrow\quad \alpha^4 < 1 $$

Step 5 · Solve $\alpha^4 < 1$

Since $\alpha^2 \ge 0$, $(\alpha^2)^2<1 \iff \alpha^2<1 \iff |\alpha|<1$, i.e.:

$$ -1 < \alpha < 1 $$

Step 6 · Apply the domain restriction

Since $z_2$ requires $\alpha\neq 0$, the solution set is:

$$ \boxed{\alpha \in (-1,0)\cup(0,1)} $$

Step 7 · Check with examples

$\alpha=\tfrac12$: $|z_1|=\sqrt{1+\tfrac14}=\tfrac{\sqrt5}{2}\approx1.118$, $|z_2|=\sqrt{1+4}=\sqrt5\approx2.236$. Indeed $|z_1|<|z_2|$. ✓

$\alpha=2$: $|z_1|=\sqrt5\approx2.236$, $|z_2|=\tfrac{\sqrt5}{2}\approx1.118$. Here $|z_1|>|z_2|$, confirming $\alpha=2$ is excluded. ✗

$\alpha=-\tfrac12$: $|z_1|=\tfrac{\sqrt5}{2}\approx1.118$, $|z_2|=\sqrt5\approx2.236$. Again $|z_1|<|z_2|$. ✓

Step 8 · Boundary and limiting behavior

At $\alpha=\pm1$: $|z_1|=|z_2|=\sqrt2$ — the two distances are equal, so $\alpha=\pm1$ are exactly the boundary points (excluded, since the inequality is strict).

As $\alpha\to0^{\pm}$: $|z_1|\to1$ while $|z_2|\to\infty$, so near the excluded point $\alpha=0$ the gap between the two distances grows without bound.