# Sum .Find the sum of z1+z2. z1=i/(1+i) z2=i/(1-i)

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### 2 Answers

We have to find the sum of z1=i/(1+i) and z2=i/(1-i)

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

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

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

=> 2i/(1 + 1)

=> i

**The sum of z1 and z2 is 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