If a + bi = (1+i)/(1-2i), find the values of the constants a and b.

The conjugate of $1-2i$ is $1+2i$.

$$a + bi = \frac{1+i}{1-2i} \times \frac{1+2i}{1+2i}$$

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

Since $i^2 = -1$:

$$= 1 + 3i + 2(-1)$$ $$= 1 + 3i - 2$$ $$= -1 + 3i$$

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

This is in the form $(x-y)(x+y) = x^2 - y^2$.

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

Since $i^2 = -1$:

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

$$a + bi = \frac{-1 + 3i}{5}$$ $$a + bi = -\frac{1}{5} + \frac{3}{5}i$$

By comparing the real and imaginary parts, we get:

$$a = -\frac{1}{5}$$ $$b = \frac{3}{5}$$

The correct option is C.