Find all the complex roots.

Find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth. The complex sixth roots of 64

The Correct Answer and Explanation is:

To find the complex sixth roots of 646464, we can represent 646464 in polar form and use De Moivre’s Theorem.

Step 1: Express 64 in Polar Form

First, write 646464 in polar form. 646464 is a real number, so its polar form is:64=64⋅(cos⁡0∘+isin⁡0∘)64 = 64 \cdot (\cos 0^\circ + i \sin 0^\circ)64=64⋅(cos0∘+isin0∘)

Here, the modulus r=64r = 64r=64, and the argument θ=0∘\theta = 0^\circθ=0∘.

Step 2: Apply De Moivre’s Theorem

De Moivre’s Theorem states that the nnn-th roots of a complex number r(cos⁡θ+isin⁡θ)r(\cos \theta + i \sin \theta)r(cosθ+isinθ) are given by:Roots=rn[cos⁡(θ+360kn)+isin⁡(θ+360kn)]\text{Roots} = \sqrt[n]{r} \left[ \cos\left(\frac{\theta + 360k}{n}\right) + i \sin\left(\frac{\theta + 360k}{n}\right) \right]Roots=nr​[cos(nθ+360k​)+isin(nθ+360k​)]

where k=0,1,2,…,n−1k = 0, 1, 2, \dots, n-1k=0,1,2,…,n−1.

In our case, r=64r = 64r=64 and n=6n = 6n=6, so the sixth roots of 646464 are:646[cos⁡(0+360k6)+isin⁡(0+360k6)]\sqrt[6]{64} \left[ \cos\left(\frac{0 + 360k}{6}\right) + i \sin\left(\frac{0 + 360k}{6}\right) \right]664​[cos(60+360k​)+isin(60+360k​)]

Since 646=2\sqrt[6]{64} = 2664​=2, the formula becomes:2[cos⁡(360k6)+isin⁡(360k6)]2 \left[ \cos\left(\frac{360k}{6}\right) + i \sin\left(\frac{360k}{6}\right) \right]2[cos(6360k​)+isin(6360k​)]

Step 3: Calculate the Roots

Now, substitute different values of kkk (from 0 to 5) into the equation.

• For k=0k = 0k=0: 2[cos⁡(0∘)+isin⁡(0∘)]=2×(1+0i)=22 \left[ \cos(0^\circ) + i \sin(0^\circ) \right] = 2 \times (1 + 0i) = 22[cos(0∘)+isin(0∘)]=2×(1+0i)=2
• For k=1k = 1k=1: 2[cos⁡(60∘)+isin⁡(60∘)]=2[12+i32]=1+i32 \left[ \cos(60^\circ) + i \sin(60^\circ) \right] = 2 \left[ \frac{1}{2} + i \frac{\sqrt{3}}{2} \right] = 1 + i\sqrt{3}2[cos(60∘)+isin(60∘)]=2[21​+i23​​]=1+i3​
• For k=2k = 2k=2: 2[cos⁡(120∘)+isin⁡(120∘)]=2[−12+i32]=−1+i32 \left[ \cos(120^\circ) + i \sin(120^\circ) \right] = 2 \left[ -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right] = -1 + i\sqrt{3}2[cos(120∘)+isin(120∘)]=2[−21​+i23​​]=−1+i3​
• For k=3k = 3k=3: 2[cos⁡(180∘)+isin⁡(180∘)]=2×(−1+0i)=−22 \left[ \cos(180^\circ) + i \sin(180^\circ) \right] = 2 \times (-1 + 0i) = -22[cos(180∘)+isin(180∘)]=2×(−1+0i)=−2
• For k=4k = 4k=4: 2[cos⁡(240∘)+isin⁡(240∘)]=2[−12−i32]=−1−i32 \left[ \cos(240^\circ) + i \sin(240^\circ) \right] = 2 \left[ -\frac{1}{2} – i \frac{\sqrt{3}}{2} \right] = -1 – i\sqrt{3}2[cos(240∘)+isin(240∘)]=2[−21​−i23​​]=−1−i3​
• For k=5k = 5k=5: 2[cos⁡(300∘)+isin⁡(300∘)]=2[12−i32]=1−i32 \left[ \cos(300^\circ) + i \sin(300^\circ) \right] = 2 \left[ \frac{1}{2} – i \frac{\sqrt{3}}{2} \right] = 1 – i\sqrt{3}2[cos(300∘)+isin(300∘)]=2[21​−i23​​]=1−i3​

Step 4: List the Roots

The six complex roots of 646464 are:

1. 222
2. 1+i31 + i\sqrt{3}1+i3​
3. −1+i3-1 + i\sqrt{3}−1+i3​
4. −2-2−2
5. −1−i3-1 – i\sqrt{3}−1−i3​
6. 1−i31 – i\sqrt{3}1−i3​

These are the complex sixth roots of 646464 in rectangular form.

Conclusion:

The six complex roots of 646464 are:2,1+i3,−1+i3,−2,−1−i3,1−i32, \quad 1 + i\sqrt{3}, \quad -1 + i\sqrt{3}, \quad -2, \quad -1 – i\sqrt{3}, \quad 1 – i\sqrt{3}2,1+i3​,−1+i3​,−2,−1−i3​,1−i3​