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