How do you simplify $(\frac{1}{1 + i}) + (\frac{1}{2 + i}) + (\frac{1}{3 + i})$ $(1/(1+i)) + (1/(2+i)) + (1/(3+i))$ ?

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How do you simplify $(\frac{1}{1 + i}) + (\frac{1}{2 + i}) + (\frac{1}{3 + i})$ $(1/(1+i)) + (1/(2+i)) + (1/(3+i))$ ?

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Answer 1 · 2016-03-19T20:06:40

$\frac{1}{1 + i} + \frac{1}{2 + i} + \frac{1}{3 + i} = \frac{6}{5} - \frac{4}{5} i$ $1/(1+i)+1/(2+i)+1/(3+i) =6/5-4/5i$

Explanation:

First, note that for any $n$ $n$ , we have

$\frac{1}{n + i} = \frac{n - i}{(n + i) (n - i)} = \frac{n - i}{n^{2} + 1}$ $1/(n+i) = (n-i)/((n+i)(n-i))=(n-i)/(n^2+1)$

If we substitute in $n = 1, 2, 3$ $n=1, 2, 3$ we get

$\frac{1}{1 + i} + \frac{1}{2 + i} + \frac{1}{3 + i} = \frac{1 - i}{2} + \frac{2 - i}{5} + \frac{3 - i}{10}$ $1/(1+i)+1/(2+i)+1/(3+i) = (1-i)/2+(2-i)/5+(3-i)/10$

$= \frac{5 (1 - i)}{10} + \frac{2 (2 - i)}{10} + \frac{3 - i}{10}$ $=(5(1-i))/10 + (2(2-i))/10 + (3-i)/10$

$= \frac{5 - 5 i + 4 - 2 i + 3 - i}{10}$ $=(5-5i+4-2i+3-i)/10$

$= \frac{12 - 8 i}{10}$ $=(12-8i)/10$

$= \frac{12}{10} - \frac{8}{10} i$ $=12/10 - 8/10i$

$= \frac{6}{5} - \frac{4}{5} i$ $=6/5-4/5i$

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How do you simplify $(\frac{1}{1 + i}) + (\frac{1}{2 + i}) + (\frac{1}{3 + i})$ $(1/(1+i)) + (1/(2+i)) + (1/(3+i))$ ?

1 Answer

Explanation:

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