Question

\( (\frac{1+i}{1-i})^{228}\)

• A. \( -4(\frac{1-i}{1+i})^{226} \)
• B. \( 4(\frac{1-i}{1+i})^{226} \)
• C. \( (\frac{1-i}{1+i})^{228} \)
• D. \( -(\frac{1-i}{1+i})^{228} \)

Hint:

Memorize the standard simplifications: \(\frac{1+i}{1-i} = i\) and \(\frac{1-i}{1+i} = -i\). These frequently appear in complex number problems and save significant calculation time.

Answer 1

The Correct Option is C

First, let's simplify the base of the expression in the question.
\( \frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i \).
So the expression becomes \( i^{228} \).
The powers of \(i\) cycle every 4 terms (\(i^1=i, i^2=-1, i^3=-i, i^4=1\)).
To evaluate \(i^{228}\), we find the remainder of 228 when divided by 4.
\(228 \div 4 = 57\) with a remainder of 0.
Therefore, \( i^{228} = i^4 = 1 \).
Now, let's simplify the base of the expression in option (C).
\( \frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i \).
The expression in option (C) is \( (-i)^{228} \).
\( (-i)^{228} = ((-1) \cdot i)^{228} = (-1)^{228} \cdot i^{228} \).
Since 228 is an even number, \((-1)^{228} = 1\).
We already calculated \(i^{228} = 1\).
So, option (C) evaluates to \(1 \cdot 1 = 1\).
Since the original expression and the expression in option (C) both equal 1, they are equivalent.