Question

Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1,$ and with $|z| = 2.$ In the complex plane, the four points $0, z, z^2,$ and $z^3$ are the vertices of a quadrilateral with area $15.$ What is the imaginary part of $z?$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

Solution:

For vertices $0, z, z^2, z^3$ the shoelace formula gives area $\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| = \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| = \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z).$

With $|z| = 2,$ this is $\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z).$ Setting $10\operatorname{Im}(z) = 15$ gives $\operatorname{Im}(z) = \dfrac32.$ (Then $\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1,$ as required.)

Thus, the correct answer is D.

2024 AMC 12B Problem 12

Solution:

Problem 12 in Other Years