# Verify if the sum is a real number 1/(1+i) + 1/(1-i)

hala718 | Certified Educator

Let S = 1/(1+i)  + 1/(1-i)

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

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

= 2/(1-i^2)

We know that i^2 = sqrt-1*sqrt-1 = -1

==> S = 2/(1-(-1)= 2/2=1

==> S = 1   which is a real number.

neela | Student

We  make the denominator real by multiplying both numerator  and denominator by the conhjugate of the demominator in each of the terms.

1/(1+i) = (1-i)/(1+i)(1-i) = (1-i)/(1- -1) = (1/2)(1+i), as i^2 = -1.

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

Therefore .

1/(1+i) +1/(1-i)

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

= (1/2){1+i+1-i} = (1/2)(2) =1.

So the  sum 1/(1+i) +1/(1-i) is 1, which is real.

giorgiana1976 | Student

For the beginning, before adding the 2 ratios, we have to transform the denominator into a real number, instead of complex numbers.

We'll multiply the complex number from denominator by it's conjugate.

If the complex number is z=a+b*i, it's conjugate is z'=a-b*i.

So, if the complex number is 1+i, it's conjugate is 1-i.

We'll multiply the first ratio by the conjugate number (1-i) and the second ratio by (1+i).

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

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

We'll eliminate like terms:

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

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

So, the sum is a real number!

thewriter | Student

### =1

As the number only has a real part and no imaginary component i.e. i, it is a real number.