How do you simplify #(5-i)/(5+i)#?

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Answer 1 · 2017-01-05T18:09:24

#12/13-5/13i#

You can multiply both the terms of the fraction by

#5-i#

and get:

#(5-i)^2/((5+i)(5-i))#

that's:

#(25-10i+i^2)/(25-i^2)#

Then, since

#i^2=-1#, you get:

#(25-10i-1)/(25+1)#

#(24-10i)/26#

#cancel24^12/cancel26^13-cancel10^5/cancel26^13i#

#12/13-5/13i#