# Sumation of the complex numbers.i/(1-i) + i/(1+i).

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### 3 Answers

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

Opening the brackets yields:

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

Using the complex number theory yields:

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

Reducing like terms in the brackets yields:

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

We have to find the value of i/(1-i) + i/(1+i)

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

=> [i(i + 1) + i(1 - i)](1 + i)(1 - i)

=> [i^2 + i + i - i^2]/(1 - i^2)

=> [2i ]/2

=> i

**The required sum is i**

To calculate the sum of 2 complex quotients that do not have a common denominator we'll have to calculate the LCD(least common denominator) of the 2 fractions.

We notice that LCD = (1+i)(1-i)

We notice also that the product (1+i)(1-i) is the result of the difference of squares

(a-b)(a+b) = a^2 - b^2

We'll write instead of product the difference of squares, where a = 1 and b = i.

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

LCD = 1^2 - i^2

We'll write instead of i^2 = -1

LCD = 1 - (-1)

LCD = 2

Now, we'll multiply the first ratio by (1-i) and the second ratio by (1+i):

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

We'll remove the brackets:

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

We'll eliminate like terms:

2i/2

We'll simplify:

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

The result of the sum of 2 quotients is a complex number, whose real part is 0 and imaginary part is 1