Complex Number Simplification
The question is to show that \( \left( \frac{1 + i \sqrt{3}}{1 - i \sqrt{3}} \right)^2 = 2 \).
We will begin by simplifying the expression inside the parentheses before squaring it.
\( \frac{1 + i \sqrt{3}}{1 - i \sqrt{3}} \)
To simplify, we can multiply both the numerator and the denominator by the conjugate of the denominator:
\( \frac{(1 + i \sqrt{3})(1 + i \sqrt{3})}{(1 - i \sqrt{3})(1 + i \sqrt{3})} \)
Expanding the numerator and the denominator:
Numerator:
\( (1 + i \sqrt{3})(1 + i \sqrt{3}) = 1 + i \sqrt{3} + i \sqrt{3} + (i \sqrt{3})^2 \)
\( = 1 + 2i \sqrt{3} - 3 \) [since \( i^2 = -1 \)]
\( = -2 + 2i \sqrt{3} \)
Denominator:
\( (1 - i \sqrt{3})(1 + i \sqrt{3}) = 1 - (i \sqrt{3})^2 \)
\( = 1 - (-3) \)
\( = 1 + 3 \)
\( = 4 \)
Now we have:
\( \frac{-2 + 2i \sqrt{3}}{4} \)
We can simplify this further by dividing both terms in the numerator by the denominator:
\( \frac{-2}{4} + \frac{2i \sqrt{3}}{4} \)
\( = -\frac{1}{2} + \frac{i \sqrt{3}}{2} \)
Now we square this result to find the original expression's square:
\( \left( -\frac{1}{2} + \frac{i \sqrt{3}}{2} \right)^2 \)
Multiplying this out gives us:
\( = \left( -\frac{1}{2} \right)^2 + 2 \cdot \left( -\frac{1}{2} \right) \cdot \left( \frac{i \sqrt{3}}{2} \right) + \left( \frac{i \sqrt{3}}{2} \right)^2 \)
\( = \frac{1}{4} - \frac{i \sqrt{3}}{2} + \frac{-3i^2}{4} \)
\( = \frac{1}{4} - \frac{i \sqrt{3}}{2} + \frac{3}{4} \) [since \( i^2 = -1 \)]
\( = 1 - \frac{i \sqrt{3}}{2} \)
However, the imaginary term does not cancel out, which makes the final expression complex. It appears there may be a mistake either in the original question or in the simplification process, as the expected result is supposed to be \( 2 \), a real number, but the working out has resulted in a complex number. It is possible that there is a typo in the question or additional constraints that were not included in the image provided.
