Question

If $z = \frac{\sqrt{3} + i}{2}$, then $z^{101} + z^{103} =$

• A. $-\sqrt{3}$
• B. $\sqrt{3}$
• C. $-i\sqrt{3}$
• D. $i\sqrt{3}$

Hint:

Any expression $z^{n-1} + z^{n+1}$ can be factored as $z^n(z + 1/z)$. Since $z + z^{-1} = 2\text{Re}(z) = \sqrt{3}$ and $z^{102} = (z^6)^{17} = (-1)^{17} = -1$, the result is simply $-\sqrt{3}$.

Answer 1

The Correct Option is A

Step 1: Switch to polar form.
Big powers are easy once we write $z$ as an angle. Here $z = \frac{\sqrt3 + i}{2}$ has size 1 and matches $\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$, so $z = e^{i\pi/6}$.

Step 2: Pull out a common power.
Factor the middle power: \[ z^{101} + z^{103} = z^{102}\left(z^{-1} + z\right) \]

Step 3: Find $z^{102}$.
Multiply the angle by 102: \[ z^{102} = e^{i \cdot 102\pi/6} = e^{i 17\pi} \]

Step 4: Simplify that angle.
Since $17\pi$ is an odd multiple of $\pi$, $e^{i17\pi} = \cos(17\pi) = -1$.

Step 5: Find $z + z^{-1}$.
Adding a number and its conjugate doubles the real part: \[ z + z^{-1} = 2\cos\frac{\pi}{6} = 2 \cdot \frac{\sqrt3}{2} = \sqrt3 \]

Step 6: Multiply.
Combine the two results: \[ z^{101} + z^{103} = (-1)(\sqrt3) = -\sqrt3 \] \[ \boxed{ -\sqrt3 } \]