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

Solution 3 (Geometry)

We can write the given condition as $\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.$ Letting $u = \left(z+\frac{1}{2}\right)^2$ , the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$ . Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$ . Let $v = z+\frac{1}{2}$ , and since the $+\frac{1}{2}$ does not change $z$ 's imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$ .

Note that the point on $\omega$ with largest magnitude is $19/4$ and argument $\pi$ (The leftmost point on $\omega$ ). The value $v$ with positive imaginary part which correspond to this point has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$ .

Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and this value has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must be the largest value.

This can non-rigorously be seen by sketching the conic figure which is the locus of $v$ .

This gives $19 + 2 \implies \textbf{(B) 21}$

~AtharvNaphade

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

Solution 3 (Geometry)

See Also