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

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

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Find trigonometric ratios using multiple identities

Full solution

Q. Find the argument of the complex number

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 $\theta = \text{arctan}(\frac{b}{a})$ . Here, $a = 3$ and $b = -\sqrt{3}$ .
Use special triangles: Calculate the arctan of $b/a$ : $\theta = \arctan(-\sqrt{3}/3)$ .
Adjust for quadrant: Using the special triangles, we know that $\arctan(-\sqrt{3}/3)$ corresponds to $-\pi/6$ because $\tan(\pi/6) = \sqrt{3}/3$ , and since $b$ is negative, the angle is in the fourth quadrant.
Find positive angle: However, the argument of a complex number is typically given as a positive angle between $0$ and $2\pi$ . To find this, we add $2\pi$ to $-\pi/6$ to get the positive angle.
Calculate final argument: Adding $2\pi$ to $-\pi/6$ gives us the final argument: $\theta = 2\pi - \pi/6 = (12\pi/6) - (1\pi/6) = 11\pi/6$ .

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