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

`\theta=251.6^\circ`

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=(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=(-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}`