Problem 76 - Algebra and Trigonometry

Question

Perform the indicated operations and write the result in standard form. $$ \frac{1+i}{1+2 i}+\frac{1-i}{1-2 i} $$

Step-by-Step Solution

Answer

The solution is \(-0.5\) or \(-\frac{1}{2}\)

Step 1: Multiply By Conjugate

Multiply both the numerator and the denominator of each fraction by the conjugate of the denominator. The conjugate of a complex number, \(a + bi\) is \(a - bi\). Accordingly, the conjugates of \(1 + 2i\) and \(1 - 2i\) are \(1 - 2i\) and \(1 + 2i\) respectively. \[\frac{1+i}{1+2 i} × \frac{1-2i}{1-2 i} + \frac{1-i}{1-2 i} × \frac{1+2i}{1+2 i} \]

Step 2: Apply Conjugate Multiplication

Apply conjugate multiplication and write the resulting complex numbers \[\frac{(1+i)(1-2i)}{(1+2 i)(1-2 i)} + \frac{(1-i)(1+2i)}{(1-2 i)(1+2 i)} = \frac{1 - 2i + i - 2 }{-4} + \frac{1 + 2i - i - 2 }{-4} = \frac{-1 - i}{-4} + \frac{-1 + i}{-4}\]

Step 3: Add the Fractions

Add the two fractions and simplify, giving the final solution in standard form \[-\frac{1}{4} +\frac{i}{4} -\frac{1}{4} -\frac{i}{4}= -\frac{1}{2}\]