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Find the argument of the complex number
9-3sqrt3i in the interval
0 <= theta < 2pi.
Express your answer in terms of
pi.
Answer:

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

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Simplify radical expressions with variables

Full solution

Q. Find the argument of the complex number

9-3 \sqrt{3} i

in the interval

0 \leq \theta<2 \pi

.

\newline

Express your answer in terms of

\pi

.

\newline

Answer:

Calculate arctan value: 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 $\theta = \text{arctan}(\frac{b}{a})$ . The complex number given is $9 - 3\sqrt{3}i$ , so $a = 9$ and $b = -3\sqrt{3}$ .
Simplify arctan expression: First, we calculate the arctan of $b/a$ , which is $\text{arctan}(-3\sqrt{3}/9)$ . Simplifying the fraction gives us $\text{arctan}(-\sqrt{3}/3)$ .
Adjust for desired range: The value of $\arctan(-\sqrt{3}/3)$ is known to be $-\pi/6$ because $\tan(-\pi/6) = -\sqrt{3}/3$ . However, since we want the argument in the interval $0 \leq \theta < 2\pi$ , we need to add $2\pi$ to $-\pi/6$ to find the equivalent angle in the desired range.
Final argument calculation: Adding $2\pi$ to $-\pi/6$ gives us $2\pi - \pi/6$ , which simplifies to $(12\pi/6) - (\pi/6) = 11\pi/6$ . Therefore, the argument of the complex number $9-3\sqrt{3}i$ in the interval $0 \leq \theta < 2\pi$ is $11\pi/6$ .

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