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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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Find the roots of factored polynomials

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:

Use formula for argument: 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 $\theta = \arctan(\frac{b}{a})$ . Here, $a = 9$ and $b = 3\sqrt{3}$ .
Calculate arctan function: Calculate the argument using the arctan function: $\theta = \arctan(\frac{3\sqrt{3}}{9})$ .
Simplify fraction: Simplify the fraction inside the arctan function: $\theta = \arctan(\sqrt{3}/3)$ .
Recognize angle in triangle: Recognize that $\arctan(\sqrt{3}/3)$ corresponds to the angle whose tangent is $\sqrt{3}/3$ . This is a well-known angle in a $30$ - $60$ - $90$ right triangle, where the angle opposite the side with length $\sqrt{3}$ is $60$ degrees or $\pi/3$ radians.
Determine final argument: Since the complex number is in the first quadrant (both $a$ and $b$ are positive), the argument is directly the $\arctan$ value: $\theta = \frac{\pi}{3}$ .

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