Question

Which of the following is the correct representation of the given expression?
\[ \frac{(1+i)^8}{(1-i)^6} \]

• A. \(2e^{-i\pi/2}\)
• B. \(-1\)
• C. \(2e^{i\pi/2}\)
• D. \(1\)

Hint:

Convert 1+i and 1-i to polar/exponential form, raise to the given powers, then simplify the ratio.

Answer 1

The Correct Option is A

Step 1: Write 1 + i and 1 - i in polar form:
The modulus of \(1+i\) is \(|1+i| = \sqrt{1^2+1^2} = \sqrt{2}\), and its argument is \(\pi/4\). So \(1+i = \sqrt{2}\, e^{i\pi/4}\).
The modulus of \(1-i\) is \(|1-i| = \sqrt{2}\), and its argument is \(-\pi/4\). So \(1-i = \sqrt{2}\, e^{-i\pi/4}\).
Step 2: Raise each to the required power:
\((1+i)^8 = (\sqrt{2})^8 e^{i \cdot 8 \cdot \pi/4} = 16\, e^{i2\pi} = 16(\cos 2\pi + i\sin 2\pi) = 16\).
\((1-i)^6 = (\sqrt{2})^6 e^{-i \cdot 6 \cdot \pi/4} = 8\, e^{-i3\pi/2}\).
Step 3: Simplify the exponential term for the denominator:
\(e^{-i3\pi/2} = \cos(-3\pi/2) + i\sin(-3\pi/2) = 0 + i(1) = i\), so \((1-i)^6 = 8i\).
Step 4: Divide the two results:
\[ \frac{(1+i)^8}{(1-i)^6} = \frac{16}{8i} = \frac{2}{i} = \frac{2}{i}\cdot\frac{-i}{-i} = \frac{-2i}{1} = -2i \]
Step 5: Express the answer in exponential form and match with options:
\(-2i = 2(\cos(-\pi/2) + i\sin(-\pi/2)) = 2e^{-i\pi/2}\), which matches option (A).
Final Answer:
\[ \boxed{2e^{-i\pi/2}} \]