2023 AMC 12A Problem 16

Problem 16 of 25IntermediateAlgebra

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?

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Solution

Write

z=x+yi.

Then

1+z+z^2

=(1+x+x^2-y^2)

+y(1+2x)i,

and the constraint is

\begin{gathered} (1+x+x^2-y^2)^2\\ {}+y^2(1+2x)^2=16. \end{gathered}

At a point where

y

is maximal on this closed, bounded curve, implicit differentiation (or Lagrange multipliers) gives

(1+2x)\bigl(P+2y^2\bigr)=0,

where

P=1+x+x^2-y^2.

But

P+2y^2=x^2+x+1+y^2\gt 0,

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,

m+n=21.

Thus, the correct answer is B.