Answer :
To determine which of the given expressions is equivalent to \(\frac{i}{1+i} - \frac{1}{2-i}\), we perform the following steps to simplify the expression:
1. Simplify \(\frac{i}{1+i}\):
Rationalize the denominator:
[tex]\[ \frac{i}{1+i} \cdot \frac{1-i}{1-i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i - i^2}{1 - i^2} \][/tex]
Simplify \(i^2 = -1\):
[tex]\[ \frac{i - (-1)}{1 - (-1)} = \frac{i + 1}{2} \][/tex]
So,
[tex]\[ \frac{i}{1+i} = \frac{1 + i}{2} \][/tex]
2. Simplify \(\frac{1}{2-i}\):
Rationalize the denominator:
[tex]\[ \frac{1}{2-i} \cdot \frac{2+i}{2+i} = \frac{1(2+i)}{(2-i)(2+i)} = \frac{2+i}{4 - (-1)} \][/tex]
Simplify \(i^2 = -1\):
[tex]\[ \frac{2+i}{5} \][/tex]
So,
[tex]\[ \frac{1}{2-i} = \frac{2+i}{5} \][/tex]
3. Subtract the two simplified fractions:
[tex]\[ \frac{1+i}{2} - \frac{2+i}{5} \][/tex]
Find a common denominator:
[tex]\[ \text{Common denominator: } 10 \][/tex]
Convert each fraction:
[tex]\[ \frac{1+i}{2} = \frac{5(1+i)}{10} = \frac{5 + 5i}{10} \][/tex]
[tex]\[ \frac{2+i}{5} = \frac{2(2+i)}{10} = \frac{4 + 2i}{10} \][/tex]
Subtract the fractions:
[tex]\[ \frac{5 + 5i}{10} - \frac{4 + 2i}{10} = \frac{(5 + 5i) - (4 + 2i)}{10} = \frac{5 + 5i - 4 - 2i}{10} = \frac{1 + 3i}{10} \][/tex]
So,
[tex]\[ \frac{i}{1+i} - \frac{1}{2-i} = \frac{1 + 3i}{10} \][/tex]
Thus, the expression equivalent to \(\frac{i}{1+i} - \frac{1}{2-i}\) is:
[tex]\[ \boxed{\frac{3i+1}{10}} \][/tex]
1. Simplify \(\frac{i}{1+i}\):
Rationalize the denominator:
[tex]\[ \frac{i}{1+i} \cdot \frac{1-i}{1-i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{i - i^2}{1 - i^2} \][/tex]
Simplify \(i^2 = -1\):
[tex]\[ \frac{i - (-1)}{1 - (-1)} = \frac{i + 1}{2} \][/tex]
So,
[tex]\[ \frac{i}{1+i} = \frac{1 + i}{2} \][/tex]
2. Simplify \(\frac{1}{2-i}\):
Rationalize the denominator:
[tex]\[ \frac{1}{2-i} \cdot \frac{2+i}{2+i} = \frac{1(2+i)}{(2-i)(2+i)} = \frac{2+i}{4 - (-1)} \][/tex]
Simplify \(i^2 = -1\):
[tex]\[ \frac{2+i}{5} \][/tex]
So,
[tex]\[ \frac{1}{2-i} = \frac{2+i}{5} \][/tex]
3. Subtract the two simplified fractions:
[tex]\[ \frac{1+i}{2} - \frac{2+i}{5} \][/tex]
Find a common denominator:
[tex]\[ \text{Common denominator: } 10 \][/tex]
Convert each fraction:
[tex]\[ \frac{1+i}{2} = \frac{5(1+i)}{10} = \frac{5 + 5i}{10} \][/tex]
[tex]\[ \frac{2+i}{5} = \frac{2(2+i)}{10} = \frac{4 + 2i}{10} \][/tex]
Subtract the fractions:
[tex]\[ \frac{5 + 5i}{10} - \frac{4 + 2i}{10} = \frac{(5 + 5i) - (4 + 2i)}{10} = \frac{5 + 5i - 4 - 2i}{10} = \frac{1 + 3i}{10} \][/tex]
So,
[tex]\[ \frac{i}{1+i} - \frac{1}{2-i} = \frac{1 + 3i}{10} \][/tex]
Thus, the expression equivalent to \(\frac{i}{1+i} - \frac{1}{2-i}\) is:
[tex]\[ \boxed{\frac{3i+1}{10}} \][/tex]
