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:

Identify parts of complex number: 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 $\theta = \text{arctan}(\frac{b}{a})$ . However, since the $\text{arctan}$ function only gives values from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ , we need to consider the quadrant in which the complex number lies to find the correct argument in the interval $0 \leq \theta < 2\pi$ .
Calculate principal value of argument: First, identify the real part $a$ and the imaginary part $b$ of the complex number $-3\sqrt{3} + 9i$ . Here, $a = -3\sqrt{3}$ and $b = 9$ .
Consider quadrant for correct argument: Next, calculate the $\arctan(\frac{b}{a})$ to find the principal value of the argument. However, since $a$ is negative and $b$ is positive, the complex number lies in the second quadrant. In the second quadrant, the argument $\theta$ is $\pi - \arctan(\left|\frac{b}{a}\right|)$ .
Compute absolute value of $b/a$ : Compute the absolute value of $b/a$ : $|\frac{b}{a}| = |\frac{9}{(-3\sqrt{3})}| = |\frac{-3}{\sqrt{3}}| = |-\sqrt{3}|$ .
Calculate $\arctan(-\sqrt{3})$ : Now, calculate $\arctan(\lvert-\sqrt{3}\rvert)$ . Since $\arctan(\sqrt{3})$ is known to be $\frac{\pi}{3}$ , $\arctan(-\sqrt{3})$ will also be $-\frac{\pi}{3}$ .
Determine argument in second quadrant: Since the complex number is in the second quadrant, the argument $\theta$ is $\pi - (-\pi/3)$ , which simplifies to $\pi + \pi/3$ .
Find argument in interval $0$ to $2\pi$ : Add $\pi$ and $\pi/3$ to find the argument of the complex number in the interval $0 \leq \theta < 2\pi$ : $\theta = \pi + \pi/3 = (3\pi/3) + (\pi/3) = 4\pi/3$ .