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: Concept
Complex numbers with large exponents are easiest to compute when converted into polar/exponential form ($z = r e^{i\theta}$) using Euler's formula.

Step 2: Meaning
The given complex number $z = \frac{\sqrt{3} + i}{2}$ can be written as: \[ z = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6} \]

Step 3: Analysis
Factor out $z^{102}$ from the given expression: \[ z^{101} + z^{103} = z^{102}(z^{-1} + z) \] Let's find $z^{102}$ and $(z + z^{-1})$: \[ z^{102} = (e^{i\pi/6})^{102} = e^{i \frac{102\pi}{6}} = e^{i 17\pi} = \cos(17\pi) + i\sin(17\pi) = -1 \] \[ z + z^{-1} = e^{i\pi/6} + e^{-i\pi/6} = 2 \cos\left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \] Multiplying these together yields: \[ z^{101} + z^{103} = (-1)(\sqrt{3}) = -\sqrt{3} \]

Step 4: Conclusion
The value of the complex expression is the real number $-\sqrt{3}$. Final Answer: (A)