# What is the sum of z1+z2=? z1=i/(1+i) z2=i/(1-i)

*print*Print*list*Cite

### 2 Answers

z1 = i/(1+i)

z2 = i/(1-i).

We make the common denominator the LCM = (1+i)(1-i) = 1- i^2 = 1-(-1) = 2.

Therefore z1 = i(1-i)/2 = i-i^2 =( i-(-1))/2 = (i+1)/2 .

z2 = i((1+i)/2 = (i+i^2)/2 = i-1)/2.

Therefore z1+z2 = {i+1 +i-1}/2 = 2i/2 = i.

Therefore z1+z2 = i.

We have to determine the result of the sum of 2 ratios.

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

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

We notice also that the product (1+i)(1-i) is like:

(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 and we'll get:

**z1 + z2 =** ** i(1-i)/2 + i(1+i)/ 2**

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

**The result is a complex number, whose real part is 0 and imaginary part is 1**