Let z be a complex number such that |z-6|=5 and |z+2-6i|=5.

Question

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:

Accepted Answer

$|z-6|=5$ is a circle centered at $(6,0)$ radius 5; $|z-(-2+6i)|=5$ is a circle centered at $(-2,6)$ radius 5.

Cartesian forms:
$$(x-6)^2+y^2=25 \implies x^2+y^2-12x+11=0 \quad(i)$$
$$(x+2)^2+(y-6)^2=25 \implies x^2+y^2+4x-12y+15=0\quad(ii)$$

Subtracting (i) from (ii): $16x-12y+4=0 \implies y=\dfrac{4x+1}{3}$.

Substituting into (i):
$$(x-6)^2+\left(\frac{4x+1}{3}\right)^2=25 \implies 9(x^2-12x+36)+16x^2+8x+1=225$$
$$25x^2-100x+100=0 \implies x^2-4x+4=0 \implies (x-2)^2=0 \implies x=2.$$

Then $y=\dfrac{4(2)+1}{3}=3$, so $z=2+3i$ (the circles are tangent, giving a single intersection point).

Now evaluate $z^3+3z^2-15z+141$ with $z=2+3i$:
$z^2=(2+3i)^2=4+12i-9=-5+12i$
$$z^3=z\cdot z^2=(2+3i)(-5+12i)=-10+24i-15i-36=-46+9i$$

$$z^3+3z^2-15z+141 = (-46+9i)+(-15+36i)+(-30-45i)+141.$$

Real: $-46-15-30+141=50$. Imaginary: $9+36-45=0$.

$z^3+3z^2-15z+141=50.$

Solutions · 1