Find the fifth roots of 243

Find the fifth roots of 243.

The correct answer and explanation is:

The problem is to find the fifth roots of 243. This means finding all complex numbers zz such that: z5=243z^5 = 243

Step 1: Express 243 in exponential form

243 is a positive real number. It can be written as a power of 3: 243=35243 = 3^5

In polar (or complex exponential) form, a positive real number rr can be written as: r=r⋅(cos⁡0+isin⁡0)=rei⋅0r = r \cdot (\cos 0 + i \sin 0) = r e^{i \cdot 0}

So, 243=35=243ei⋅0243 = 3^5 = 243 e^{i \cdot 0}

Step 2: Use De Moivre’s theorem for roots

The general formula for the nnth roots of a complex number reiθr e^{i \theta} is: zk=r1/nei(θ+2πkn)fork=0,1,2,…,n−1z_k = r^{1/n} e^{i \left(\frac{\theta + 2\pi k}{n}\right)} \quad \text{for} \quad k = 0, 1, 2, …, n-1

Here, n=5n = 5, r=243r = 243, and θ=0\theta = 0.

Calculate the magnitude of the roots: r1/5=2431/5=(35)1/5=3r^{1/5} = 243^{1/5} = (3^5)^{1/5} = 3

Step 3: Calculate the fifth roots

Using the formula: zk=3ei0+2πk5=3(cos⁡2πk5+isin⁡2πk5)z_k = 3 e^{i \frac{0 + 2\pi k}{5}} = 3 \left(\cos \frac{2\pi k}{5} + i \sin \frac{2\pi k}{5}\right)

for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

The roots are:

• For k=0k=0:

z0=3(cos⁡0+isin⁡0)=3(1+0i)=3z_0 = 3(\cos 0 + i \sin 0) = 3(1 + 0i) = 3

• For k=1k=1:

z1=3(cos⁡2π5+isin⁡2π5)z_1 = 3 \left(\cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5}\right)

• For k=2k=2:

z2=3(cos⁡4π5+isin⁡4π5)z_2 = 3 \left(\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5}\right)

• For k=3k=3:

z3=3(cos⁡6π5+isin⁡6π5)z_3 = 3 \left(\cos \frac{6\pi}{5} + i \sin \frac{6\pi}{5}\right)

• For k=4k=4:

z4=3(cos⁡8π5+isin⁡8π5)z_4 = 3 \left(\cos \frac{8\pi}{5} + i \sin \frac{8\pi}{5}\right)

Step 4: Approximate values

Using approximate values of cosine and sine:

• cos⁡2π5≈0.3090\cos \frac{2\pi}{5} \approx 0.3090, sin⁡2π5≈0.9511\sin \frac{2\pi}{5} \approx 0.9511
• cos⁡4π5≈−0.8090\cos \frac{4\pi}{5} \approx -0.8090, sin⁡4π5≈0.5878\sin \frac{4\pi}{5} \approx 0.5878
• cos⁡6π5≈−0.8090\cos \frac{6\pi}{5} \approx -0.8090, sin⁡6π5≈−0.5878\sin \frac{6\pi}{5} \approx -0.5878
• cos⁡8π5≈0.3090\cos \frac{8\pi}{5} \approx 0.3090, sin⁡8π5≈−0.9511\sin \frac{8\pi}{5} \approx -0.9511

So:

• z0=3z_0 = 3
• z1=3(0.3090+0.9511i)=0.927+2.853iz_1 = 3(0.3090 + 0.9511i) = 0.927 + 2.853i
• z2=3(−0.8090+0.5878i)=−2.427+1.763iz_2 = 3(-0.8090 + 0.5878i) = -2.427 + 1.763i
• z3=3(−0.8090−0.5878i)=−2.427−1.763iz_3 = 3(-0.8090 – 0.5878i) = -2.427 – 1.763i
• z4=3(0.3090−0.9511i)=0.927−2.853iz_4 = 3(0.3090 – 0.9511i) = 0.927 – 2.853i

Explanation

The number 243 is a perfect fifth power because it equals 353^5. Taking the fifth root gives a magnitude of 3. Because complex numbers have multiple roots equally spaced around the unit circle, there are five distinct fifth roots spaced evenly in angle by 2π5\frac{2\pi}{5} radians. This is why the roots have the same magnitude but differ in their argument (angle). The principal root is the real positive root 33, while the other four are complex conjugates and symmetrically placed in the complex plane. This method uses De Moivre’s theorem, which connects the algebraic roots of complex numbers to trigonometric forms.

This explains why every complex number has exactly nn distinct nnth roots and shows the structure of these roots geometrically on the complex plane.