🧩 Problem & Options

Find \(k\) if \(3^{49}(x+iy)=\left(\tfrac{3}{2}+i\tfrac{\sqrt3}{2}\right)^{100}\) and \(x=ky\).

Options: (a) \(-\tfrac13\)   (b) \(\sqrt3\)   (c) \(-\sqrt3\)   (d) \(-\tfrac{1}{\sqrt3}\)

✅ Final Answer

\( \boxed{k=-\tfrac{1}{\sqrt3}} \)   (Option d)

🧠 Solution — Polar Form + De Moivre

1) Extract the unit-modulus factor

\(\displaystyle \frac{3}{2}+i\frac{\sqrt3}{2} =3\Big(\frac{\sqrt3}{2}+i\frac12\Big) =3\big(\cos\tfrac{\pi}{6}+i\sin\tfrac{\pi}{6}\big).\)

\[ \left(\tfrac{3}{2}+i\tfrac{\sqrt3}{2}\right)^{100} =3^{100}\!\left(\cos\tfrac{\pi}{6}+i\sin\tfrac{\pi}{6}\right)^{100} =3^{100}\!\left(\cos\tfrac{50\pi}{3}+i\sin\tfrac{50\pi}{3}\right). \]

Since \(\tfrac{50\pi}{3}=16\pi+\tfrac{2\pi}{3}\), we get \(\cos\tfrac{50\pi}{3}+i\sin\tfrac{50\pi}{3} =-\tfrac12+i\tfrac{\sqrt3}{2}\).

2) Equate both sides and use \(x=ky\)

\[ 3^{49}(k+ i)y = 3^{100}\!\left(-\tfrac12 + i\tfrac{\sqrt3}{2}\right) \;\Rightarrow\; (k+i)y = 3\!\left(-\tfrac12 + i\tfrac{\sqrt3}{2}\right). \]

3) Compare real/imaginary parts

\[ \text{Re: } ky=-\tfrac{3}{2}, \qquad \text{Im: } y=\tfrac{3\sqrt3}{2} \;\Rightarrow\; k=\frac{-\tfrac{3}{2}}{\tfrac{3\sqrt3}{2}}=-\frac{1}{\sqrt3}. \]

⚠️ Pitfalls

• Forgetting to reduce \( \tfrac{50\pi}{3} \) modulo \(2\pi\) to \( \tfrac{2\pi}{3} \).
• Not substituting \(x=ky\) early; it makes equating parts trivial.
• Confusing \( \tfrac{3}{2}+i\tfrac{\sqrt3}{2} \) with the unit vector; factor the \(3\) first.

📝 Quick Check