Question

$(\sqrt{3}+i)^{10} + (\sqrt{3}-i)^{10} =$

• A. $1024\sqrt{3}$
• B. 1024
• C. 2048
• D. $512\sqrt{3}$

Hint:

For expressions of the form $(x+iy)^n + (x-iy)^n$, using De Moivre's theorem is the most efficient method. Remember that the sum will be $2r^n\cos(n\theta)$ and the difference will be $2ir^n\sin(n\theta)$.

Answer 1

The Correct Option is B

Let $z = \sqrt{3}+i$. The expression is $z^{10} + \bar{z}^{10}$.
We know that for any complex number $z$, $z^n + \bar{z}^n = 2 \text{Re}(z^n)$.
First, we convert $z$ to its polar form, $z = r(\cos\theta + i\sin\theta)$.
The modulus is $r = |z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$.
The argument is $\theta = \arctan(\frac{1}{\sqrt{3}})$, which is $\frac{\pi}{6}$ since the point is in the first quadrant.
So, $z = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))$.
Now, we use De Moivre's theorem to find $z^{10}$:
$z^{10} = 2^{10}(\cos(10 \cdot \frac{\pi}{6}) + i\sin(10 \cdot \frac{\pi}{6}))$.
$z^{10} = 1024(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}))$.
The real part of $z^{10}$ is $1024 \cos(\frac{5\pi}{3})$.
We can evaluate $\cos(\frac{5\pi}{3}) = \cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
So, $\text{Re}(z^{10}) = 1024 \times \frac{1}{2} = 512$.
The required sum is $2 \text{Re}(z^{10}) = 2 \times 512 = 1024$.