Question

The value of $\dfrac{(1+i)^n}{(1-i)^{n-4}}$, where $i=\sqrt{-1}$ and $n$ is an integer, is:

• A. $\dfrac{i^n}{4}$
• B. $4i^n$
• C. $-4i^n$
• D. $-1$
• E. $1$

Hint:

Use $\frac{1+i}{1-i} = i$ and $(1-i)^4 = -4$ to simplify quickly.

Answer 1

The Correct Option is C

Concept:
• Convert complex numbers into polar/exponential or use identities
• $(1+i)(1-i) = 2$

Step 1: Rewrite expression
\[ \frac{(1+i)^n}{(1-i)^{n-4}} = (1+i)^n \cdot (1-i)^{4-n} \]

Step 2: Group terms
\[ = \frac{(1+i)^n}{(1-i)^n} \cdot (1-i)^4 \]

Step 3: Simplify ratio
\[ \frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1+2i+i^2}{2} = \frac{2i}{2} = i \] \[ \Rightarrow \left(\frac{1+i}{1-i}\right)^n = i^n \]

Step 4: Compute $(1-i)^4$
\[ (1-i)^2 = 1 - 2i + i^2 = -2i \] \[ (1-i)^4 = (-2i)^2 = 4i^2 = -4 \]

Step 5: Final multiplication
\[ = i^n \cdot (-4) = -4i^n \] Final Conclusion:
\[ = -4i^n \]