Evaluate (1+i)⁻¹ and express the result in the form a+bi.

The conjugate of $(1+i)$ is $(1-i)$.

$$\frac{1}{1+i} \times \frac{1-i}{1-i}$$

The numerator becomes $1 \times (1-i) = 1-i$.
The denominator is a product of a complex number and its conjugate, which simplifies to $a^2 + b^2$ for $(a+bi)(a-bi)$. Here, $a=1$ and $b=1$.

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

So the expression becomes:

$$\frac{1-i}{2}$$

$$\frac{1}{2} - \frac{1}{2}i$$

Comparing this with the given options, it matches option (c).

The final answer is $\text{(c) } \left(\frac{1}{2} - \frac{1}{2}i\right)$.