Question

(√3+I)^20= please solve this with using Moivre's formula

Answer 1

(√3+i)^20 can be divided using the Moivre formula:

(√3+i)^20 = (3^1/2+i)^(20) = (3^(1/2))^(20) + I*(3^(1/2)(19)) + ( I^2)^(20-319) = 3^10 + i*3^5 + 1^2

Thus, (√3 + i)^20 = 531.

Answer 2

Certainly! To solve the expression \((\sqrt{3} + i)^{20}\) using Moivre's formula, we'll follow these steps:

Moivre's formula states that for any complex number \(z = r(\cos \theta + i \sin \theta)\), the nth power of \(z\) can be found using the formula:

\[ z^n = r^n (\cos n\theta + i \sin n\theta) \]

In the given expression, \(\sqrt{3} + i\) can be represented in polar form as \(2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})\) since \(\sqrt{3}\) and \(i\) form a right-angled triangle with a 30-degree angle.

Now, let's use Moivre's formula to find \((\sqrt{3} + i)^{20}\):

\[ r = 2 \] \[ \theta = \frac{\pi}{6} \]

The formula for \(z^n\) becomes:

\[ (2^n) (\cos n\theta + i \sin n\theta) \]

Substitute \(n = 20\):

\[ (2^{20}) (\cos \frac{20\pi}{6} + i \sin \frac{20\pi}{6}) \]

Simplify the angle:

\[ (2^{20}) (\cos \frac{10\pi}{3} + i \sin \frac{10\pi}{3}) \]

Now, we know that \(\cos \frac{10\pi}{3} = \cos \frac{4\pi}{3}\) and \(\sin \frac{10\pi}{3} = -\sin \frac{4\pi}{3}\), so the expression becomes:

\[ (2^{20}) (\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) \]

Finally, evaluate the cosine and sine values:

\[ (2^{20}) \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) \]

Multiply this by \(2^{20}\) to get the final result:

\[ -2^{20} - i 2^{20} \sqrt{3} \]

So, \((\sqrt{3} + i)^{20} = -2^{20} - i 2^{20} \sqrt{3}\).