How do you simplify #(1/(1+i)) + (1/(2+i)) + (1/(3+i)) #?

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Answer 1 · 2016-03-19T20:06:40

#1/(1+i)+1/(2+i)+1/(3+i) =6/5-4/5i#

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

#1/(n+i) = (n-i)/((n+i)(n-i))=(n-i)/(n^2+1)#

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

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

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

#=(5-5i+4-2i+3-i)/10#

#=(12-8i)/10#

#=12/10 - 8/10i#

#=6/5-4/5i#