Question

The complex number \( 3-3 \) i in trogonometric form is: a. 23 cis \( 330^{\circ} \) b. 23 cis \( 30^{\circ} \) c. 23 cis \( 60^{\circ} \) d. 23 cis \( 300^{\circ} \)

Answer 1

The complex number 3 - 3i can be written in the form a + bi, where a is the real part and b is the imaginary part. In this case, a = 3 and b = -3.

To convert a complex number from rectangular form (a + bi) to trigonometric form (r cis θ), we can use the following formulas:

r = |a + bi| = sqrt(a^2 + b^2)
θ = arctan(b/a) + kπ, where k is an integer and the angle is measured in radians.

In this case, we have:

r = sqrt(3^2 + (-3)^2) = sqrt(18) = 3sqrt(2)
θ = arctan((-3)/3) + kπ = -π/4 + kπ, where k is an integer.

To find the principal argument, we use k = 0:

θ = -π/4

Therefore, the complex number 3 - 3i in trigonometric form is:

3sqrt(2) cis (-π/4)

Converting this to degrees, we get:

3sqrt(2) cis (-45°)

So the answer is not one of the options given.