1. We are asked to find the square root of the complex number $\sqrt{2} - i$ and express it in the form $a + bi$ where $a$ and $b$ are real numbers. 2. Recall that for a complex number $z = x + yi$, its square root can be written as $\sqrt{r}(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2})$, where $r = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \arg(z)$ is the argument (angle) of $z$. 3. First, find the modulus $r$: $$r = \sqrt{(\sqrt{2})^2 + (-1)^2} = \sqrt{2 + 1} = \sqrt{3}$$ 4. Next, find the argument $\theta$: Since $x = \sqrt{2} > 0$ and $y = -1 < 0$, the point lies in the fourth quadrant. $$\theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{-1}{\sqrt{2}}\right) = -\arctan\left(\frac{1}{\sqrt{2}}\right)$$ 5. Calculate $\arctan\left(\frac{1}{\sqrt{2}}\right)$: This is approximately $0.61548$ radians. So, $$\theta \approx -0.61548$$ 6. The square root has modulus $\sqrt{r} = \sqrt{\sqrt{3}} = 3^{1/4}$. 7. The argument of the square root is $\frac{\theta}{2} \approx \frac{-0.61548}{2} = -0.30774$ radians. 8. To get the solution with the smallest positive angle, add $2\pi$ to the negative angle: $$-0.30774 + 2\pi \approx 5.97544$$ But since $-0.30774$ is already the principal value, the smallest positive angle solution is the one with angle $2\pi - 0.30774 = 5.97544$ radians. 9. Calculate $a$ and $b$: $$a = 3^{1/4} \cos(-0.30774) = 3^{1/4} \cos(0.30774)$$ $$b = 3^{1/4} \sin(-0.30774) = -3^{1/4} \sin(0.30774)$$ 10. Numerically, $$3^{1/4} = \sqrt{\sqrt{3}} \approx 1.3161$$ $$\cos(0.30774) \approx 0.9535$$ $$\sin(0.30774) \approx 0.3023$$ 11. Therefore, $$a \approx 1.3161 \times 0.9535 = 1.255$$ $$b \approx -1.3161 \times 0.3023 = -0.398$$ 12. The square root of $\sqrt{2} - i$ in $a + bi$ form with the smallest positive angle is approximately: $$1.255 - 0.398i$$