find the trigonometric form of -12-12(square root)3i
a. 24(cos2pi/3 + isin 2pi/3)
b. 24(cos4pi/3 + isin 4pi/3)
c. 12(cos 4pi/3 + isin 4pi/3)
d. 12(cos12pi/3 + isin 2pi/3)
e. 12(squareroot) 2 (cos4pi/3 + isin 4pi/3)
-12-12√3 i
magnitude = √( 144 + 432)
= √576 = 24
tan Ø = -12√3/-12) = √3
Ø = 60° or π/3
but from my sketch I see that we are in quadrant III
so Ø = π/3 + π = 4π/3
-12-12√3 i
= 24(cos 4π/3 + i sin 4π/3) or 24 cis 4π/3
looks like b)
To find the trigonometric form of a complex number, we can use the formula:
z = r(cosθ + isinθ),
where r is the magnitude (distance from the origin) of the complex number, and θ is the argument (angle) of the complex number with the positive real axis.
In this case, the complex number given is -12 - 12√3i. To find the magnitude (r):
r = √((-12)^2 + (-12√3)^2)
= √(144 + 432)
= √576
= 24.
To find the argument (θ), we can use the following relationship:
tanθ = Im/Re,
where Im is the imaginary part and Re is the real part of the complex number.
In this case:
tanθ = (-12√3) / (-12)
= √3.
To determine the value of θ, we find the reference angle by taking the arctan of the magnitude:
θ = arctan(√3)
= π/3.
Since the given complex number is in the third quadrant, we adjust the angle by adding π, giving:
θ = π + π/3
= 4π/3.
Putting it all together, we have:
z = 24(cos(4π/3) + isin(4π/3)).
Comparing this with the given options, we see that the correct choice is:
b. 24(cos(4π/3) + isin(4π/3)).