Question

Consider the following. Cube roots of −343 (a) Use the formula zk = n r cos θ + 2πk n + i sin θ + 2πk n to find the indicated roots of the complex number. (Enter your answers in trigonometric form. Let 0 ≤ θ < 2π.) z0 = z1 = z2 =

Answer 1

Answer:

Z0 = 7 ( cos 60° + isin60°)

Z1 = 7( cos180° + isin180° )

Z2 = 7 ( cos300° + isin300°)

Step-by-step explanation:

Given that;

cube root of  -343

so, z = -343 + 0i

∴ r = √ (( -343)² + (0)²)  = 343

so tan∅ = y/x ⇒ tan∅ = 0/-343

∅ = tan⁻¹ (0/-343)

= 0 or 180°

but we are going yo make use of 180° since -343 is negative x-axis

Zk = ∛343 ( cos 180/3 + 360K/3) + isin(180/3 + 360k/3)

here k = 0, 1, 2, 3 .........

SO z0 = z1 = z2 = ???

k=0

Z0 =  ∛343 ( cos 180/3 + isin180/3)

=  ∛343 ( cos 60° + isin60°)

Z0 = 7 ( cos 60° + isin60°)

K=1

Z1 = ∛343 ( cos 180/3 + 360×1 / 3 + isin180/3 + 360×1 / 3 )

= ∛343 ( cos180° + isin180°)

Z1 = 7( cos180° + isin180° )

K=2

Z2 = ∛343 ( cos 180/3 + 360×2 / 3 + isin180/3 + 360×2 / 3 )

= ∛343 ( cos300° + isiN300°)

Z2 = 7 ( cos300° + isin300°)