Question

Find the argument of the complex number -3–9 in the interval 0°< ϴ < 360°,
rounding to the nearest tenth of a degree if necessary.

Answer 1

Answer:

$\theta=251.6^\circ$

Step-by-step explanation:

Complex Numbers

They are expressed as the sum of a real part and an imaginary part:

$Z = a+b\mathbf{ i}$

Complex numbers can also be expressed in polar form:

$Z = r(\cos\theta+\sin\theta \mathbf{ i}) = r. cis(\theta)$

Where r is the modulus of the complex number and θ is the argument.

The argument can be calculated by:

$\displaystyle \tan\theta=\frac{b}{a}$

The angle θ must be calculated in the appropriate quadrant depending on the signs of the real and imaginary parts.

The complex number is given as:

$Z = -3 -9\mathbf{ i}$

Here: a=-3, b=-9

Since both components are negative, the argument lies in the third quadrant (180° < θ < 270°).

$\displaystyle \tan\theta=\frac{-9}{-3}$

$\displaystyle \tan\theta=3$

$\theta=\arctan(3)$

The calculator gives the answer 71.6°, we need to adjust the angle to the third quadrant by adding 180°, thus

$\mathbf{\theta=251.6^\circ}$