10.50 Among the complex numbers satisfying the condition $|z-5 i| \leq 4$, find the one whose argument has the smallest positive value.
10.50 The inequality $|z-5 i| \leq 4$ is satisfied by points in the plane lying inside and on the boundary of a circle with center $M(0 ; 5)$ and radius 4. From Fig. 10.15, it is clear that the number with the smallest positive argument corresponds to the point $K$, where the line drawn from the origin is tangent to the circle.

Draw $K P \perp O M$. In the right triangle $O K M$, we have $O K=\sqrt{5^{2}-4^{2}}=3$; further, since

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-337.jpg?height=552&width=352&top_left_y=828&top_left_x=801)

Fig. 10.15 $O K^{2}=O P \cdot O M$, then $O P=\frac{O K^{2}}{O M}=1.8$, and $K P=\sqrt{3^{2}-1.8^{2}}=2.4$.

Thus, the desired number is $2.4+1.8 i$.

Answer: $2.4+1.8 i$.