Question - Complex Number Simplification
mathcomplex number simplificationimaginary number divisionirrational number handlingmathematical expression simplification
Solution:
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.
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