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

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Answer 1 · 2018-06-22T22:50:01

#(1-i)/(2)#

#(1)/(1+i)#

You multiply by the complex conjugate the denominator with the sign changed:

#(1-i)/(1-i)#

#(1)/(1+i)*(1-i)/(1-i)#

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

#(1-i)/(1-i^2)#

#(1-i)/(1-sqrt(-1)^2)#

#(1-i)/(1-(-1))#

#(1-i)/(2)#

Answer 2 · 2018-06-23T02:45:21

#color(maroon)(=> (1 - i)/2#

#1 / (1 + i)#

#=> (1 / 91 + i) * ((1 - i) / (1 - i)), " rationalizing the denominator, by multiplying & Dividing by the conjugate of the denominator"#

#=> (1 - i) / (1 - i^2)#

#=> (1/2) - i/2, " as " i^2 = -1#