Let z be a complex number such that |z-6|=5 a | vedclass

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(C) The given equations represent two circles in the complex plane:$|z-6|=5$ represents a circle with center $C_{1}(6, 0)$ and radius $r_{1}=5$.$|z-(-

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$50$

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(C) The given equations represent two circles in the complex plane:$|z-6|=5$ represents a circle with center $C_{1}(6, 0)$ and radius $r_{1}=5$.$|z-(-2+6i)|=5$ represents a circle with center $C_{2}(-2, 6)$ and radius $r_{2}=5$.The distance between the centers is $C_{1}C_{2} = \sqrt{(6 - (-2))^2 + (0 - 6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10$.Since $C_{1}C_{2} = r_{1} + r_{2} = 5 + 5 = 10$,the two circles touch each other externally at a single point $z$.This point $z$ is the midpoint of the line segment connecting the centers $C_{1}$ and $C_{2}$.$z = \frac{(6, 0) + (-2, 6)}{2} = (2, 3)$,which corresponds to $z = 2 + 3i$.Now,we evaluate the expression $z^{3}+3z^{2}-15z+141$ at $z = 2+3i$.$(z-2) = 3i \implies (z-2)^2 = (3i)^2 = -9 \implies z^2 - 4z + 4 = -9 \implies z^2 = 4z - 13$.Multiply by $z$: $z^3 = 4z^2 - 13z = 4(4z - 13) - 13z = 16z - 52 - 13z = 3z - 52$.Substitute these into the expression: $(3z - 52) + 3(4z - 13) - 15z + 141 = 3z - 52 + 12z - 39 - 15z + 141 = (3+12-15)z + (-52-39+141) = 0z + 50 = 50$.

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

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