The expression provided is:

$\frac{4 + 5i}{1 + i}$

To simplify this, we can multiply both the numerator and the denominator by the conjugate of the denominator, $1 - i$, in order to eliminate the imaginary unit $i$ from the denominator.

Step-by-Step Process:

1. Multiply numerator and denominator by the conjugate:

   $\frac{(4 + 5i)}{(1 + i)} \times \frac{(1 - i)}{(1 - i)} = \frac{(4 + 5i)(1 - i)}{(1 + i)(1 - i)}$

2. Simplify the denominator:

   $(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$

   So the denominator is $2$.

3. Expand the numerator:

   $(4 + 5i)(1 - i) = 4(1 - i) + 5i(1 - i)$

   Now distribute:

   $4(1 - i) = 4 - 4i$ $5i(1 - i) = 5i - 5i^2$

   Recall that $i^2 = -1$, so:

   $5i - 5i^2 = 5i - 5(-1) = 5i + 5 = 5 + 5i$

4. Combine like terms:

   $(4 - 4i) + (5 + 5i) = (4 + 5) + (-4i + 5i) = 9 + i$

5. Final expression:

   $\frac{9 + i}{2}$

   This can be written as:

   $\frac{9}{2} + \frac{1}{2}i$

So, the simplified form of the expression is:

$\boxed{\frac{9}{2} + \frac{1}{2}i}$

Would you like more details or have any questions?

Related Questions:

1. What is the conjugate of a complex number, and how is it used in division?
2. How can you verify if your simplified result is correct?
3. What happens when you multiply complex numbers?
4. Can this technique be applied to any complex fraction division?
5. How would the solution change if the denominator was $1 - i$ instead?

Tip:

When simplifying complex fractions, multiplying by the conjugate of the denominator always helps to remove the imaginary part from the denominator.