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Chapter 1: Problem 76

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

Short Answer

Expert verified
The result in standard form is \( 1.2 \).

Step by step solution

01

Simplify the denominators

To simplify the denominators, multiply both the numerator and denominator of each fraction by the conjugate of the denominator. This will result in a real number. The conjugate of a complex number is obtained by changing the sign in front of the imaginary part. Hence, \(\frac{1+i}{1+2 i}\) is multiplied by \(\frac{1-2 i}{1-2 i}\) and \(\frac{1-i}{1-2 i}\) is multiplied by \(\frac{1+2 i}{1+2 i}\).
02

Calculate

That results in: \(\frac{(1+i)(1-2 i)}{(1+2 i)(1-2 i)} + \frac{(1-i)(1+2 i)}{(1-2 i)(1+2 i)}\). This simplifies to: \(\frac{1 - 2i + i - 2i^2}{1 - 4i^2} + \frac{1 + 2i - i - 2i^2}{1 - 4i^2}\).
03

Simplify further

Recall that \(i^2 = -1\). Thus, \(\frac{1 - 2i + i + 2}{1 + 4} + \frac{1 + 2i - i + 2}{1 + 4}\),which simplifies to: \(\frac{3 - i}{5} + \frac{3 + i}{5}\).
04

Final result

Finally, add the two fractions together: \(\frac{3 - i}{5} + \frac{3 + i}{5} = \frac{3+3}{5} = \frac{6}{5} = 1.2\)

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