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Find the argument of the complex number
-2sqrt2+2sqrt2i in the interval
0 <= theta < 2pi. Express your answer in terms of
pi.
Answer:

Find the argument of the complex number $-2 \sqrt{2}+2 \sqrt{2} i$ in the interval $0 \leq \theta<2 \pi$ . 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

-2 \sqrt{2}+2 \sqrt{2} i

in the interval

0 \leq \theta<2 \pi

. Express your answer in terms of

\pi

.

\newline

Answer:

Identify Quadrant: The complex number given is $-2\sqrt{2} + 2\sqrt{2}i$ . Here, $a = -2\sqrt{2}$ and $b = 2\sqrt{2}$ . Both $a$ and $b$ are negative and positive respectively, which places the complex number in the second quadrant.
Calculate $|\frac{b}{a}|$ : In the second quadrant, the argument $\theta$ is $\pi - \arctan(|\frac{b}{a}|)$ . Let's calculate $|\frac{b}{a}|$ : $\newline$ |\frac{b}{a}| = |\frac{\(2\)\sqrt{\(2\)}}{\(-2\)\sqrt{\(2\)}}| = |\(-1| = $1$
Calculate $\arctan(1)$ : Now we calculate $\arctan(1)$ , which is known to be $\frac{\pi}{4}$ or $45$ degrees.
Calculate Argument Theta: Since the complex number is in the second quadrant, the argument theta is: $\theta = \pi - \arctan(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

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