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: Simplify $(1+i)^2$ and $(1-i)^2$ first:
$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$.
$(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.
Step 2: Build up $(1+i)^8$ using repeated squaring:
$(1+i)^4 = [(1+i)^2]^2 = (2i)^2 = 4i^2 = -4$.
$(1+i)^8 = [(1+i)^4]^2 = (-4)^2 = 16$.
Step 3: Build up $(1-i)^6$ using repeated squaring:
$(1-i)^4 = [(1-i)^2]^2 = (-2i)^2 = 4i^2 = -4$.
$(1-i)^6 = (1-i)^4 \cdot (1-i)^2 = (-4)(-2i) = 8i$.
Step 4: Form the ratio and rationalize:
$$\frac{(1+i)^8}{(1-i)^6} = \frac{16}{8i} = \frac{2}{i}$$Multiply numerator and denominator by $i$: $\frac{2}{i}\cdot\frac{i}{i} = \frac{2i}{i^2} = \frac{2i}{-1} = -2i$.
Step 5: Convert to exponential (polar) form to match the given options:
The number $-2i$ has modulus $2$ and lies on the negative imaginary axis, so its argument is $-\pi/2$. Thus $-2i = 2e^{-i\pi/2}$, confirming option (A).
Final Answer:
\[ \boxed{2e^{-i\pi/2}} \]