Complex Numbers – Geometry of Complex Numbers – JEE Main 28 Jan 2026 Shift 1

Solution:

The given equations represent two circles:
$C_1: |z – 6| = 5 \Rightarrow$ Center $O_1(6, 0)$, Radius $r_1 = 5$
$C_2: |z – (-2 + 6i)| = 5 \Rightarrow$ Center $O_2(-2, 6)$, Radius $r_2 = 5$

Calculate the distance between the centers $O_1$ and $O_2$:
$$d = \sqrt{(6 – (-2))^2 + (0 – 6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10$$

Since $d = r_1 + r_2 = 5 + 5 = 10$, the two circles touch each other externally at a single point.

The point of contact $z$ is the midpoint of the line segment joining $O_1$ and $O_2$:
$$z = \frac{6 + (-2)}{2} + i\left(\frac{0 + 6}{2}\right) = \frac{4}{2} + 3i = 2 + 3i$$

Now we need to find the value of $z^3 + 3z^2 – 15z + 141$.
Since $z = 2 + 3i$, we have:
$$z – 2 = 3i$$
Squaring both sides:
$$(z – 2)^2 = (3i)^2$$
$$z^2 – 4z + 4 = -9$$
$$z^2 – 4z + 13 = 0$$

We divide the polynomial $z^3 + 3z^2 – 15z + 141$ by $z^2 – 4z + 13$:
$$z^3 + 3z^2 – 15z + 141 = z(z^2 – 4z + 13) + 7z^2 – 28z + 141$$
Since $z^2 – 4z + 13 = 0$, the first term becomes 0.
$$= 7z^2 – 28z + 141$$
$$= 7(z^2 – 4z) + 141$$
From $z^2 – 4z = -13$, substitute this value:
$$= 7(-13) + 141$$
$$= -91 + 141$$
$$= 50$$

Ans. (3)