How do you simplify (1/(1+i)) + (1/(2+i)) + (1/(3+i)) (11+i)+(12+i)+(13+i)?

1 Answer
Mar 19, 2016

1/(1+i)+1/(2+i)+1/(3+i) =6/5-4/5i11+i+12+i+13+i=6545i

Explanation:

First, note that for any nn, we have

1/(n+i) = (n-i)/((n+i)(n-i))=(n-i)/(n^2+1)1n+i=ni(n+i)(ni)=nin2+1

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

1/(1+i)+1/(2+i)+1/(3+i) = (1-i)/2+(2-i)/5+(3-i)/1011+i+12+i+13+i=1i2+2i5+3i10

=(5(1-i))/10 + (2(2-i))/10 + (3-i)/10=5(1i)10+2(2i)10+3i10

=(5-5i+4-2i+3-i)/10=55i+42i+3i10

=(12-8i)/10=128i10

=12/10 - 8/10i=1210810i

=6/5-4/5i=6545i