Find the fifth root of 32i that graphs in the first quadrant: 2 ](cos[ Jo + i sin[ 19) Use degree measure:
The Correct Answer and Explanation is:
Correct Answer:
The fifth root of 32i32i32i that lies in the first quadrant is: 2(cos(18∘)+isin(18∘))2 \left( \cos\left(18^\circ\right) + i \sin\left(18^\circ\right) \right)2(cos(18∘)+isin(18∘))
Explanation:
To find the fifth roots of a complex number like 32i32i32i, the number must first be written in polar form. A complex number z=r(cosθ+isinθ)z = r(\cos \theta + i \sin \theta)z=r(cosθ+isinθ) can be rooted using De Moivre’s Theorem: z1/n=r1/n(cos(θ+360∘kn)+isin(θ+360∘kn)),k=0,1,…,n−1z^{1/n} = r^{1/n} \left( \cos\left(\frac{\theta + 360^\circ k}{n} \right) + i \sin\left(\frac{\theta + 360^\circ k}{n} \right) \right), \quad k = 0, 1, …, n-1z1/n=r1/n(cos(nθ+360∘k)+isin(nθ+360∘k)),k=0,1,…,n−1
First, convert 32i32i32i to polar form. The modulus is: r=∣32i∣=02+322=32r = |32i| = \sqrt{0^2 + 32^2} = 32r=∣32i∣=02+322=32
The angle θ\thetaθ (argument) is the angle from the positive real axis to the vector 32i32i32i, which is located on the positive imaginary axis. So: θ=90∘\theta = 90^\circθ=90∘
Now, apply De Moivre’s Theorem to find the fifth roots: Let n=5,r1/5=321/5=2\text{Let } n = 5,\quad r^{1/5} = 32^{1/5} = 2Let n=5,r1/5=321/5=2
Now compute the five roots using: zk=2(cos(90∘+360∘k5)+isin(90∘+360∘k5))z_k = 2 \left( \cos\left( \frac{90^\circ + 360^\circ k}{5} \right) + i \sin\left( \frac{90^\circ + 360^\circ k}{5} \right) \right)zk=2(cos(590∘+360∘k)+isin(590∘+360∘k))
for k=0,1,2,3,4k = 0, 1, 2, 3, 4k=0,1,2,3,4. The angles are:
- k=0→90∘5=18∘k = 0 \rightarrow \frac{90^\circ}{5} = 18^\circk=0→590∘=18∘
- k=1→450∘5=90∘k = 1 \rightarrow \frac{450^\circ}{5} = 90^\circk=1→5450∘=90∘
- k=2→810∘5=162∘k = 2 \rightarrow \frac{810^\circ}{5} = 162^\circk=2→5810∘=162∘
- k=3→1170∘5=234∘k = 3 \rightarrow \frac{1170^\circ}{5} = 234^\circk=3→51170∘=234∘
- k=4→1530∘5=306∘k = 4 \rightarrow \frac{1530^\circ}{5} = 306^\circk=4→51530∘=306∘
Only the root at 18° lies in the first quadrant (where both cosine and sine are positive). Therefore, the correct fifth root is: 2(cos(18∘)+isin(18∘))\boxed{2 \left( \cos(18^\circ) + i \sin(18^\circ) \right)}2(cos(18∘)+isin(18∘))
