Solve the equation (z+1)/(1+i) + (z)/(1−i) = 5+5i for the complex number z.

The given equation is:

$$\frac{z+1}{1+i} + \frac{z}{1-i} = 5+5i$$

To combine the fractions, find a common denominator, which is $(1+i)(1-i)$.

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

Now rewrite the left-hand side with the common denominator:

$$\frac{(z+1)(1-i)}{(1+i)(1-i)} + \frac{z(1+i)}{(1-i)(1+i)} = 5+5i$$ $$\frac{(z+1)(1-i) + z(1+i)}{2} = 5+5i$$

Expand the terms in the numerator:

$$(z+1)(1-i) = z(1) + z(-i) + 1(1) + 1(-i) = z - zi + 1 - i$$ $$z(1+i) = z + zi$$

Substitute these back into the numerator:

$$(z - zi + 1 - i) + (z + zi)$$

Combine like terms:

$$z - zi + 1 - i + z + zi = (z+z) + (-zi+zi) + 1 - i$$ $$= 2z + 0 + 1 - i$$ $$= 2z + 1 - i$$

$$\frac{2z + 1 - i}{2} = 5+5i$$

Multiply both sides by 2:

$$2z + 1 - i = 2(5+5i)$$ $$2z + 1 - i = 10+10i$$

Isolate the term with $z$:

$$2z = 10+10i - (1-i)$$ $$2z = 10+10i - 1 + i$$

Combine the real and imaginary parts on the right-hand side:

$$2z = (10-1) + (10+1)i$$ $$2z = 9 + 11i$$

Divide by 2 to find $z$:

$$z = \frac{9+11i}{2}$$ $$z = \frac{9}{2} + \frac{11}{2}i$$

The complex number $z$ in the form $a+bi$ is $\boxed{\frac{9}{2} + \frac{11}{2}i}$.