Determine if i/(1+i) + i/ (1-i) is real or imaginary.

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

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To determine if i/ (1+i) + i/ (1-i) is real or imaginary, we have to convert the expression to make the denominator the same for both the terms. On simplification if we are left with an imaginary number the original expression is imaginary else it is real.

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

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

=> i [1-i +1+i] / (1^2 – i^2)

=> i*2 / (1+1)

=> i*2/2

=> i

Therefore we find that i/ (1+i) + i/ (1-i) can be simplified to i. Hence it is imaginary.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We have to determine the result of the sum of 2 ratios and we'll have to decide if the result is a complex or real number.

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:

 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.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

We  multiply the Numerator and denominators by the the denominator's conjugates. (a+bi abd a-bi are conjugates). Then the given expresion becomes:

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

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

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

=2i/2

= i.

Therefore the given expression i/(1+i)+i/(1-i)= i which is imaginary.

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