Complex Numbers Part 2

Concept:The equations $|z-6|=5$ and $|z+2-6i|=5$ represent circles in the complex plane with the same radius.Explanation:Let $z=x+iy$.The first circle has centre $(6,0)$ and radius $5$.Its Cartesian equation: $(x-6)^2+y^2=25$.Expanding gives $x^2+y^2-12x+11=0$.The second circle has centre $(-2,6)$ and radius $5$.Its Cartesian equation: $(x+2)^2+(y-6)^2=25$.Expanding gives $x^2+y^2+4x-12y+15=0$.Subtract the first equation from the second to find the line of intersection (common chord):$(4x+12x) -12y + (15-11)=0$ simplifies to $16x-12y+4=0$.Divide by $4$: $4x-3y+1=0$, so $y=\frac{4x+1}{3}$.Substitute $y$ into the first circle equation:$(x-6)^2 + \left(\frac{4x+1}{3}\right)^2 = 25$.Multiply by $9$: $9(x^2-12x+36) + (16x^2+8x+1) = 225$.Simplify: $9x^2-108x+324+16x^2+8x+1=225$ → $25x^2-100x+100=0$.Divide by $25$: $x^2-4x+4=0$ → $(x-2)^2=0$ → $x=2$.Then $y=\frac{4(2)+1}{3} = \frac{9}{3}=3$.Thus $z=2+3i$.Now compute the expression $z^3+3z^2-15z+141$.First, $z^2=(2+3i)^2=4+12i+9i^2=4+12i-9=-5+12i$.Then $z^3=z\cdot z^2 = (2+3i)(-5+12i) = -10+24i-15i+36i^2 = -10+9i-36 = -46+9i$.Also $3z^2 = 3(-5+12i) = -15+36i$.And $-15z = -15(2+3i) = -30-45i$.Add all terms: $(-46+9i) + (-15+36i) + (-30-45i) + 141$.Real parts: $-46-15-30+141 = 50$.Imaginary parts: $9i+36i-45i = 0$.Hence the value is $50$.Answer:The value is $50$, which corresponds to option D.