1. **Stating the problem:** We are given two complex numbers $z_1 = 3 - i$ and $z_2 = -3 + i$. We need to find their modulus and principal argument, and express them in polar form. 2. **Formula and rules:** - The modulus of a complex number $z = x + yi$ is given by $$|z| = \sqrt{x^2 + y^2}$$ - The principal argument $\theta$ is the angle the vector makes with the positive real axis, calculated by $$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$ adjusted to the correct quadrant. - The polar form of a complex number is $$z = r(\cos\theta + i\sin\theta)$$ where $r$ is the modulus and $\theta$ is the principal argument. 3. **Calculate modulus and argument for $z_1 = 3 - i$:** - Real part $x = 3$, imaginary part $y = -1$ - Modulus: $$|z_1| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$ - Argument: $$\theta_1 = \tan^{-1}\left(\frac{-1}{3}\right) = \tan^{-1}(-\frac{1}{3})$$ Since $x > 0$ and $y < 0$, $z_1$ lies in the fourth quadrant, so the principal argument is negative: $$\theta_1 \approx -0.32175 \text{ radians}$$ 4. **Calculate modulus and argument for $z_2 = -3 + i$:** - Real part $x = -3$, imaginary part $y = 1$ - Modulus: $$|z_2| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$$ - Argument: $$\theta_2 = \tan^{-1}\left(\frac{1}{-3}\right) = \tan^{-1}(-\frac{1}{3})$$ Since $x < 0$ and $y > 0$, $z_2$ lies in the second quadrant, so we add $\pi$ to the angle: $$\theta_2 = \pi + \tan^{-1}(-\frac{1}{3}) \approx 3.4633 \text{ radians}$$ 5. **Express in polar form:** - For $z_1$: $$z_1 = \sqrt{10} \left(\cos(-0.32175) + i \sin(-0.32175)\right)$$ - For $z_2$: $$z_2 = \sqrt{10} \left(\cos(3.4633) + i \sin(3.4633)\right)$$ **Final answers:** - $|z_1| = \sqrt{10}$, $\arg(z_1) \approx -0.32175$ radians - $|z_2| = \sqrt{10}$, $\arg(z_2) \approx 3.4633$ radians - Polar forms as above.