Complex Number 08F60E

1. **State the problem:** Simplify or analyze the complex number $$-1 - i\sqrt{5}$$. 2. **Recall the form of a complex number:** A complex number is generally written as $$a + bi$$ where $$a$$ and $$b$$ are real numbers and $$i$$ is the imaginary unit with $$i^2 = -1$$. 3. **Identify components:** Here, $$a = -1$$ and $$b = -\sqrt{5}$$. 4. **Magnitude (modulus) of the complex number:** The magnitude $$r$$ is given by $$r = \sqrt{a^2 + b^2}$$. 5. **Calculate magnitude:** $$r = \sqrt{(-1)^2 + (-\sqrt{5})^2} = \sqrt{1 + 5} = \sqrt{6}$$. 6. **Argument (angle) $$\theta$$:** $$\theta = \arctan\left(\frac{b}{a}\right) = \arctan\left(\frac{-\sqrt{5}}{-1}\right) = \arctan(\sqrt{5})$$. Since both $$a$$ and $$b$$ are negative, the complex number lies in the third quadrant, so add $$\pi$$ to the principal value: $$\theta = \pi + \arctan(\sqrt{5})$$. 7. **Polar form:** $$-1 - i\sqrt{5} = r(\cos\theta + i\sin\theta) = \sqrt{6}(\cos(\pi + \arctan(\sqrt{5})) + i\sin(\pi + \arctan(\sqrt{5})))$$. **Final answer:** The complex number $$-1 - i\sqrt{5}$$ has magnitude $$\sqrt{6}$$ and argument $$\pi + \arctan(\sqrt{5})$$.