Question

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

• A. 19
• B. 21
• C. 23
• D. 25

Answer 1

Answer: (C)

23

Answer 2

We start with the equation

$$\lvert 1+z+z^2\rvert = 4,$$

where $z = x+iy$ is a complex number. Our goal is to find the maximum value of the imaginary part of $z$; i.e. maximize $y$.

A very useful trick is to “complete the square” in the quadratic in $z$. Notice that

$$z^2+z+1 = \left(z+\tfrac{1}{2}\right)^2 + \tfrac{3}{4}.$$

Define

$$w = z+\tfrac{1}{2}.$$

Then the equation becomes

$$\left|w^2 + \frac{3}{4}\right| = 4.$$

In this substitution the imaginary part of $z$ is the same as that of $w$ (since $z = w - \tfrac{1}{2}$ and $-\tfrac{1}{2}$ is real). Write

$$w = u+iv.$$

Then

$$w^2 = (u+iv)^2 = u^2-v^2+2iuv,$$

so that

$$w^2 + \frac{3}{4} = \left(u^2-v^2+\frac{3}{4}\right) + 2iuv.$$

Thus the modulus condition becomes

$$\left(u^2-v^2+\frac{3}{4}\right)^2 + (2uv)^2 = 16.$$

Rather than tackling this directly in $u,v$, an alternative and simpler method is to “optimize” directly in $z=x+iy$. Write

$$1+z+z^2 = 1 + (x+iy) + \bigl(x^2-y^2+2ixy\bigr) = \Bigl(1+x+x^2-y^2\Bigr) + i\Bigl(y+2xy\Bigr).$$

Denote

$$A = 1+x+x^2-y^2,\quad B = y+2xy.$$

Thus the condition is:

$$A^2+B^2 = 16.$$

For a given $y$ we need the existence of real $x$ such that

$$\bigl(x^2+x+(1-y^2)\bigr)^2 + \bigl(y(2x+1)\bigr)^2 = 16.$$

To find the extreme (maximum) value of $y$, we require that for that $y$ the equation in $x$ has a double (repeated) solution. In other words, we consider the function:

$$F(x)=\Bigl[x^2+x+(1-y^2)\Bigr]^2 + y^2\,(2x+1)^2 - 16 = 0,$$

and we require that

$$\frac{dF}{dx}=0.$$

Differentiating with respect to $x$,

$$\frac{dF}{dx}= 2\Bigl[x^2+x+(1-y^2)\Bigr](2x+1)+ 2y^2\,(2x+1)\cdot 2.$$

Factor out the common term $2(2x+1)$:

$$\frac{dF}{dx}= 2(2x+1)\Bigl[x^2+x+(1-y^2)+2y^2\Bigr]=2(2x+1)\Bigl[x^2+x+1+y^2\Bigr].$$

Since the quadratic $x^2+x+1+y^2$ is always positive for all real $x$ and $y$ (its discriminant is negative), the only possibility for a double solution is if

$$2x+1=0\quad\Longrightarrow\quad x=-\frac{1}{2}.$$

Substitute $x=-\frac{1}{2}$ back into the expressions. First, compute the real part:

$$A=x^2+x+(1-y^2) \quad \text{with}\quad x=-\frac{1}{2}.$$

Calculate:

$$\left(-\frac{1}{2}\right)^2 -\frac{1}{2} = \frac{1}{4}-\frac{1}{2} = -\frac{1}{4},$$

so that

$$A = -\frac{1}{4} + (1-y^2) = \frac{3}{4} - y^2.$$

Next, the imaginary part:

$$B = y\,(2x+1) = y\,(2(-\frac{1}{2})+1)= y\,( -1 + 1) = 0.$$

Thus the modulus condition becomes:

$$\left(\frac{3}{4}-y^2\right)^2 = 16.$$

Taking square roots gives

$$\frac{3}{4} - y^2 = \pm 4.$$

Consider the two cases:

1. $\frac{3}{4}-y^2=4 \quad\Longrightarrow\quad -y^2 = 4 - \frac{3}{4}=\frac{13}{4},$ so $y^2 = -\frac{13}{4}$ (not possible).

2. $\frac{3}{4}-y^2=-4 \quad\Longrightarrow\quad y^2 = \frac{3}{4}+4=\frac{3}{4}+\frac{16}{4}=\frac{19}{4}.$

Thus,

$$y=\pm\frac{\sqrt{19}}{2}.$$

Since we require the maximum (largest positive) imaginary part,

$$y_{\text{max}}=\frac{\sqrt{19}}{2}.$$

The answer is given in the form $\sqrt{\frac{m}{n}}$ which matches $\sqrt{\frac{19}{4}}$ with $m=19$ and $n=4$. Therefore,

$$m+n=19+4=23.$$

Below is a mermaid diagram summarizing the key steps:

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Explanation (Minimal):

1. Write $1+z+z^2 = \bigl(x^2+x+(1-y^2)\bigr)+ i\bigl(y+2xy\bigr)$ for $z=x+iy$.
2. For a maximum $y$, the equation $\left(x^2+x+(1-y^2)\right)^2 + \left(y+2xy\right)^2=16$ must have a double root in $x$. Differentiate w.r.t. $x$ and set $\frac{dF}{dx}=0$. This leads to $2x+1=0$, i.e. $x = -\frac{1}{2}$.
3. Substitute $x=-\frac{1}{2}$ to get $\left(\frac{3}{4}-y^2\right)^2 = 16$, which gives $y^2=\frac{19}{4}$.
4. Hence, maximum $y = \frac{\sqrt{19}}{2} = \sqrt{\frac{19}{4}}$; thus $m=19,\; n=4$ and $m+n=23$.

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