2013 AMC 12A Problems/Problem 25

Problem

Let $f : \mathbb{C} \to \mathbb{C}$ be defined by $f(z) = z^2 + iz + 1$ . How many complex numbers $z$ are there such that $\text{Im}(z) > 0$ and both the real and the imaginary parts of $f(z)$ are integers with absolute value at most $10$ ?

$\textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441$

Solution

Suppose $f(z)=z^2+iz+1=c=a+bi$ . We look for $z$ with $\operatorname{Im}(z)>0$ such that $a,b$ are integers where $|a|, |b|\leq 10$ .

First, use the quadratic formula:

$z = \frac{1}{2} (-i \pm \sqrt{-1-4(1-c)}) = -\frac{i}{2} \pm \sqrt{ -\frac{5}{4} + c }$

Generally, consider the imaginary part of a radical of a complex number: $\sqrt{u}$ , where $u = v+wi = r e^{i\theta}$ .

$\operatorname{Im}(\sqrt{u}) = \operatorname{Im}(\pm \sqrt{r} e^{i\theta/2}) = \pm \sqrt{r} \sin(\theta/2) = \pm \sqrt{r}\sqrt{\frac{1-\cos\theta}{2}} = \pm \sqrt{\frac{r-v}{2}}$ .

Now let $u= -5/4 + c$ , then $v = -5/4 + a$ , $w=b$ , $r=\sqrt{v^2 + w^2}$ .

Note that $\operatorname{Im}(z)>0$ if and only if $\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}$ . The latter is true only when we take the positive sign, and that $r-v > 1/2$ ,

or $v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2$ , $w^2 > 1/4 + v$ , or $b^2 > a-1$ .

In other words, when $b^2 > a-1$ , the equation $f(z)=a+bi$ has unique solution $z$ in the region $\operatorname{Im}(z)>0$ ; and when $b^2 \leq a-1$ there is no solution. Therefore the number of desired solution $z$ is the same as the number of ordered pairs $(a,b)$ such that integers $|a|, |b|\leq 10$ , and that $b^2 \geq a$ .

When $a\leq 0$ , there is no restriction on $b$ so there are $11\cdot 21 = 231$ pairs;

when $a > 0$ , there are $2(1+4+9+10+10+10+10+10+10+10)=2(84)=168$ pairs.

So there are $231+168=\boxed{399}$ in total.

Solution 2 (motivated by coordinate geometry)

We consider the function $f(z)$ as a mapping from the 2-D complex plane onto itself. We complete the square of $f(z)=z^2+iz+1=(z+\frac{i}{2})^2+\frac{5}{4}$ .

Now, we must decide the range of $f(z)$ based on the domain of $z$ , $\operatorname{Im}(z)>0$ . To do this, we are interested in mapping the boundary line $\operatorname{Im}(z)=0$ . To make the mapping simpler, let $f(z)=g(z)+\frac{5}{4}$ , or $g(z)=(z+\frac{i}{2})^2$ .

We intend to map of the line $\operatorname{Im}(z)=0$ using the function $g(z)$ . This transformation is equivalent to the polar equation $r=(\frac{1}{2}\csc(\frac{\theta}{2}))^2$ . Using polar and trig identities, we can restate this equation as the rectangular form of a parabola,

$x=y^2-\frac{1}{4}$ ,

where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$ . So, we conclude that $f(z)$ maps the line $\operatorname{Im}(z)=0$ to the parabola

$x=y^2-\frac{1}{4}+\frac{5}{4}=y^2+1$ .

A quick check reveals that the range of $f(z)$ is to the left of the parabola, meaning that any point on or to the right of parabola cannot be reached.

Since the problem requires $|\operatorname{Re}(z)|$ and $|\operatorname{Im}(z)|$ to both be integers and at most 10, all that remains is counting all points with integer coordinates in the range of $f(z), \operatorname{Im}(z)>0$ . To do this, we employ complementary counting.

The points of interest are $|\operatorname{Re}(z)|\leq 10$ and $|\operatorname{Im}(z)|\leq 10$ , resulting in a total of $441$ points. For lattice points on or to the right of the parabola, there are $10$ points for $x=0$ , $9$ points for $x=\pm 1$ , $6$ points for $x=\pm 2$ , and $1$ point for $x=\pm 3$ . Summing it all together, our answer is $441-(10+2*9+2*6+2*1)=\boxed{399}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/365

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2013 AMC 12A Problems/Problem 25

Contents

Problem

Solution

Solution 2 (motivated by coordinate geometry)

Video Solution by Richard Rusczyk

See also