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Simplify the following expression: (1 + 1)/((1+i)ⁿ - 1)(1+i)ⁿ - 1

Mathematics

ScanSolve AI Answer

Answer

(1+i)n((1+i)n−1)2\mathbf{\frac{(1+i)^n}{((1+i)^n - 1)^2}}

Step-by-step solution

4 steps

Let's simplify this expression step by step.

  1. Identify the given expression.

    The expression from the image is:

    1+1(1+i)n−1(1+i)n−1\frac{1 + \frac{1}{(1+i)^n - 1}}{(1+i)^n - 1}
  2. Simplify the numerator of the main fraction.

    The numerator is 1+1(1+i)n−11 + \frac{1}{(1+i)^n - 1}. To combine these terms, find a common denominator, which is (1+i)n−1(1+i)^n - 1.

    1+1(1+i)n−1=(1+i)n−1(1+i)n−1+1(1+i)n−11 + \frac{1}{(1+i)^n - 1} = \frac{(1+i)^n - 1}{(1+i)^n - 1} + \frac{1}{(1+i)^n - 1}

    Combine the numerators:

    (1+i)n−1+1(1+i)n−1=(1+i)n(1+i)n−1\frac{(1+i)^n - 1 + 1}{(1+i)^n - 1} = \frac{(1+i)^n}{(1+i)^n - 1}
  3. Substitute the simplified numerator back into the main expression.

    Now the expression becomes:

    (1+i)n(1+i)n−1(1+i)n−1\frac{\frac{(1+i)^n}{(1+i)^n - 1}}{(1+i)^n - 1}
  4. Simplify the complex fraction.

    To simplify a complex fraction ABC\frac{\frac{A}{B}}{C}, we can rewrite it as AB×1C=ABC\frac{A}{B} \times \frac{1}{C} = \frac{A}{BC}.
    In this case, A=(1+i)nA = (1+i)^n, B=(1+i)n−1B = (1+i)^n - 1, and C=(1+i)n−1C = (1+i)^n - 1.

    (1+i)n(1+i)n−1×1(1+i)n−1\frac{(1+i)^n}{(1+i)^n - 1} \times \frac{1}{(1+i)^n - 1}

    Multiply the numerators and the denominators:

    (1+i)n((1+i)n−1)((1+i)n−1)\frac{(1+i)^n}{((1+i)^n - 1)((1+i)^n - 1)} (1+i)n((1+i)n−1)2\frac{(1+i)^n}{((1+i)^n - 1)^2}
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