Find the argument of the complex number $-3 \sqrt{3}-9 i$ in the interval $0 \leq \theta<2 \pi$ . Express your answer in terms of $\pi$ . $\newline$ Answer:

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Q. Find the argument of the complex number

-3 \sqrt{3}-9 i

in the interval

0 \leq \theta<2 \pi

. Express your answer in terms of

\pi

\newline

Answer:

Calculate atan $2$ ratio: To find the argument of a complex number in the form $a + bi$ , where $a$ is the real part and $b$ is the imaginary part, we use the formula $\theta = \text{atan2}(b, a)$ . The complex number given is $-3\sqrt{3} - 9i$ , so $a = -3\sqrt{3}$ and $b = -9$ .
Determine quadrant: First, we calculate the arctangent of the ratio of the imaginary part to the real part using atan $2$ , which takes into account the signs of $a$ and $b$ to determine the correct quadrant for the argument. $\theta = \text{atan2}(-9, -3\sqrt{3})$
Adjust angle range: Since both the real and imaginary parts are negative, the complex number lies in the third quadrant. The arctangent function will give us a principal value in the range $\$-\pi, \pi$ \), but we need to adjust it to be in the range $\$0, 2\pi$ \) for the final answer.
Convert to positive values: To adjust the angle to the correct range, we add $\pi$ to the principal value because the angle in the third quadrant is $\pi$ radians more than the angle in the first quadrant. $\newline$ $\theta = \text{atan2}(-9, -3\sqrt{3}) + \pi$
Simplify arctangent: Using the properties of arctangent, we know that $\text{atan2}(-9, -3\sqrt{3})$ is the same as $\text{atan2}(9, 3\sqrt{3})$ , but since we are in the third quadrant, we add $\pi$ to the result. $\newline$ $\theta = \text{atan2}(9, 3\sqrt{3}) + \pi$
Finalize argument: The arctangent of the ratio $\frac{9}{3\sqrt{3}}$ simplifies to the arctangent of $\sqrt{3}$ , which is $\frac{\pi}{3}$ . $\theta = \frac{\pi}{3} + \pi$
Finalize argument: The arctangent of the ratio $\frac{9}{3\sqrt{3}}$ simplifies to the arctangent of $\sqrt{3}$ , which is $\frac{\pi}{3}$ .
$\theta = \frac{\pi}{3} + \pi$ Adding $\pi$ to $\frac{\pi}{3}$ gives us $\frac{4\pi}{3}$ , which is the argument of the complex number in the interval $0 \leq \theta < 2\pi$ .
$\theta = \frac{4\pi}{3}$