2023 AMC 12A Problems/Problem 16

Problem

Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution 1

First, substitute in $z=a+bi$ .

$|1+(a+bi)+(a+bi)^2|=4$ $|(1+a+a^2-b^2)+ (b+2ab)i|=4$ $(1+a+a^2-b^2)^2+ (b+2ab)^2=16$ $(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16$

Let $p=b^2$ and $q=1+a+a^2$

$(q-p)^2+ p(4q-3)=16$ $p^2-2pq+q^2 + 4pq -3p=16$

We are trying to maximize $b=\sqrt p$ , so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$ .

$p^2+(2q-3)p+(q^2-16)=0$ $p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}$

We want to maximize $p$ , due to the fact that $q$ is always negatively contributing to $p$ 's value, that means we want to minimize $q$ .

Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$

If we plug $q$ 's minimum value in, we get that $p$ 's maximum value is $p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}$

Then $b=\frac{\sqrt{19}}{2}$ and $m+n=\boxed{21}$

- CherryBerry

Solution 2

We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$ . Rearranging the quadratic to standard form and applying the quadratic formula, we have $z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{-3-4c}}{2}.$ The imaginary part of $z$ is maximized when $c=4$ , making it $i\sqrt{19}/2$ . Thus the answer is $\boxed{21}$ .

~cantalon

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2023 AMC 12A Problems/Problem 16

Contents

Problem

Solution 1

Solution 2

See Also