Let's solve the given problem step-by-step:
[tex]\[ \frac{1 + 2i}{1 - i} + \frac{1 - 2i}{1 + i} \][/tex]
### Step 1: Simplify each term individually.
#### First term: [tex]\(\frac{1 + 2i}{1 - i}\)[/tex]
To simplify this expression, multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{1 + 2i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{(1 + 2i)(1 + i)}{(1 - i)(1 + i)} \][/tex]
Calculate the numerator:
[tex]\[ (1 + 2i)(1 + i) = 1(1) + 1(i) + 2i(1) + 2i(i) = 1 + i + 2i + 2i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex]:
[tex]\[ 1 + i + 2i + 2(-1) = 1 + i + 2i - 2 = -1 + 3i \][/tex]
Calculate the denominator:
[tex]\[ (1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 1 + 1 = 2 \][/tex]
Hence, the first term simplifies to:
[tex]\[ \frac{-1 + 3i}{2} = -\frac{1}{2} + \frac{3i}{2} \][/tex]
#### Second term: [tex]\(\frac{1 - 2i}{1 + i}\)[/tex]
Again, multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{1 - 2i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{(1 - 2i)(1 - i)}{(1 + i)(1 - i)} \][/tex]
Calculate the numerator:
[tex]\[ (1 - 2i)(1 - i) = 1(1) + 1(-i) - 2i(1) - 2i(-i) = 1 - i - 2i + 2i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex]:
[tex]\[ 1 - i - 2i + 2(-1) = 1 - i - 2i - 2 = -1 - 3i \][/tex]
Calculate the denominator:
[tex]\[ (1 + i)(1 - i) = 1 - i^2 = 1 - (-1) = 1 + 1 = 2 \][/tex]
Hence, the second term simplifies to:
[tex]\[ \frac{-1 - 3i}{2} = -\frac{1}{2} - \frac{3i}{2} \][/tex]
### Step 2: Add the simplified terms.
Combine the two results:
[tex]\[ -\frac{1}{2} + \frac{3i}{2} + \left(-\frac{1}{2} - \frac{3i}{2}\right) \][/tex]
Combine like terms:
[tex]\[ \left(-\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{3i}{2} - \frac{3i}{2}\right) \][/tex]
[tex]\[ -1 + 0i \][/tex]
Thus, the simplified form of the expression is:
[tex]\[ -1 + 0i \][/tex]
So, in the form [tex]\( a + ib \)[/tex], the answer is:
[tex]\[ a = -1 \quad \text{and} \quad b = 0 \][/tex]
