# Need to solve a sumi/(1-i) + i/(1+i)

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### 3 Answers

To solve i/(1-i) +i/(1+i).

i is the sqrt (-1). So i^2 = -1.

To simplify the expression, we multiply both numerator and denominator by the same expression and also we convert the denominator into a real number.

The first term:

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

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

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

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

The second term:

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

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

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

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

Adding (1) and (2) , we get:

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

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

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

We need to find the result of i/(1-i) + i/(1+i)

Now we have : i/(1-i) + i/(1+i)

let's make the denominator equal.

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

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

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

=> 2i / 1^2 - i^2

=> 2i / 1 +1

=> 2i / 2

=> i

**Therefore i/(1-i) + i/(1+i) simplified 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**