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How do you calculate #(1+i)/(1-i)#?

1 Answer

Jan 5, 2018

#i#

Explanation:

We can compute this by multiplying both numerator and denominator by the conjugate, #1+i#, of the denominator.

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

We know that #i^2=-1#, so:

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

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