Let z be a complex number such that |z - 6| = 5 and |z + 2 - 6i| = 5. Then the value of $$z^{3}+3z^{2}-15z+141$$ is equal to

We are given the complex number $$z$$ satisfying the two equations $$|z - 6| = 5$$ and $$|z + 2 - 6i| = 5$$, and we wish to find the value of the expression $$z^3 + 3z^2 - 15z + 141$$.

The equation $$|z - 6| = 5$$ represents the circle centered at $$(6,0)$$ with radius 5, while $$|z + 2 - 6i| = 5$$ represents the circle centered at $$(-2,6)$$ with the same radius. Writing $$z = x + iy$$, these become

$$ (x-6)^2 + y^2 = 25, $$

$$ (x+2)^2 + (y-6)^2 = 25. $$

Subtracting the second equation from the first gives

$$ (x-6)^2 - (x+2)^2 + y^2 - (y-6)^2 = 0, $$

which simplifies to

$$ -16x + 12y - 4 = 0, $$

or equivalently

$$ 4x - 3y + 1 = 0 \quad\Longrightarrow\quad x = \frac{3y - 1}{4}. $$

Substituting $$x = \tfrac{3y - 1}{4}$$ into $$(x-6)^2 + y^2 = 25$$ yields

$$ \Bigl(\frac{3y - 1}{4} - 6\Bigr)^2 + y^2 = 25, $$

$$ \Bigl(\frac{3y - 25}{4}\Bigr)^2 + y^2 = 25, $$

$$ \frac{9y^2 - 150y + 625}{16} + y^2 = 25, $$

$$ 9y^2 - 150y + 625 + 16y^2 = 400, $$

$$ 25y^2 - 150y + 225 = 0, $$

$$ y^2 - 6y + 9 = 0, $$

so $$(y - 3)^2 = 0$$ and hence $$y = 3$$. Substituting back gives $$x = \tfrac{3\cdot 3 - 1}{4} = 2$$, and therefore $$z = 2 + 3i$$.

Next we compute

$$ z^2 = (2+3i)^2 = 4 + 12i - 9 = -5 + 12i, $$

$$ z^3 = z\cdot z^2 = (2+3i)(-5+12i) = -10 + 24i - 15i + 36i^2 = -46 + 9i. $$

Finally,

$$ z^3 + 3z^2 - 15z + 141 = (-46+9i) + 3(-5+12i) - 15(2+3i) + 141 $$

$$ = -46 + 9i - 15 + 36i - 30 - 45i + 141 $$

$$ = (-46 - 15 - 30 + 141) + (9 + 36 - 45)i $$

$$ = 50 + 0i = 50. $$

Hence, the value of the expression is 50.