# Which is the value of the number i/(1-i)? Is the same value of the number i/(1+i)?

hala718 | Certified Educator

i/1-i = i/1+i

cross multiply:

i(1+i) = i(1-i)

==> i+i^2 = i-i^2

==> 2i^2 = =

==> i^2 = 0

==> i =0

then :

i/1+i= 0/1 =0

i/1-i= 0

neela | Student

The value of i/(1-i) is  i (1+i)/{(1-i)(1+i)} =( i+i^2)/(1-i^2) = (i-1)/2, where i = sqrt(-1) and therefore i^2 = -1.

So the value of the number is (i-1)/2.

The value of i/(1+i) = i(1-i)/{(1+i)(1-i)} = (i-i^2)/ (1-i^2) = (i+1)/2

So the given two  numbers  are not equal as their diffrence = (i+1)/2 -(i-1)/2 = 1.

giorgiana1976 | Student

It is easy to notice that both, numerator and denominator, are complex numbers. According to the rule, if we have a ratio which has at denominator a complex number, this one has to be multiplied with the conjugate of the complex number, which in this case is (1+i).

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

After opening the brackets and knowing from the theory of the complex number that i^2=-1, we'll get:

i/(1-i)=(i-1)/2

In the same way we'll calculate i/(1+i), having now as conjugate (1-i).

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

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

So, the number will be:

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

From both results we notice that: i/(1-i)=i/(1+i).