Question

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 $\dfrac{\sqrt{m}}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$20$

$21$

$22$

$23$

$24$

Solution:

Write $z=x+yi.$ Then $1+z+z^2=(1+x+x^2-y^2)+y(1+2x)i,$ and the constraint is $(1+x+x^2-y^2)^2+y^2(1+2x)^2=16.$

Setting the derivative of $y$ with respect to $x$ to zero factors as $(1+2x)\bigl(P+2y^2\bigr)=0,$ where $P=1+x+x^2-y^2.$ The factor $P+2y^2=0$ is impossible for real $x,y,$ so $x=-\tfrac12.$

Then $1+2x=0,$ so the constraint reduces to $\left(\tfrac34-y^2\right)^2=16.$ Taking $\tfrac34-y^2=-4$ gives $y^2=\tfrac{19}{4},$ so the maximum is $y=\dfrac{\sqrt{19}}{2}.$

Here $m=19$ and $n=2,$ so $m+n=21.$

Thus, the correct answer is B.

2023 AMC 12A Problem 16

Solution:

Problem 16 in Other Years