Find the argument of the complex number -3sqrt3+9i in the interval 0

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Find the argument of the complex number  -3sqrt3+9i in the interval  0 <= theta < 2pi. Express your answer in terms of  pi. Answer:

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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 = arctan(b/a). However, since the arctan function only gives values from -pi/2 to pi/2, we need to consider the quadrant in which the complex number lies to find the correct argument in the interval 0 <= theta < 2pi.Calculate principal value of argument: First, identify the real part (a) and the imaginary part (b) of the complex number -3sqrt3 + 9i. Here, a = -3sqrt3 and b = 9.Consider quadrant for correct argument: Next, calculate the arctan(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(|b/a|).Compute absolute value of b/a: Compute the absolute value of b/a: |b/a| = |9/(-3sqrt3)| = |-3/sqrt3| = |-sqrt3|.Calculate arctan(-sqrt3): Now, calculate arctan(|-sqrt3|). Since arctan(sqrt3) is known to be pi/3, arctan(-sqrt3) will also be -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 2pi: Add pi and pi/3 to find the argument of the complex number in the interval 0 <= theta < 2pi: theta = pi + pi/3 = (3pi/3) + (pi/3) = 4pi/3.

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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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: