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, for all $z$ , $f(z)=a+bi$ satisfies $b^2 > a-1$ , with a unique solution $z$ . Therefore we need to count 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 $231+168=399$ in total.

2013 AMC 12A (Problems • Answer Key • Resources)
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2013 AMC 12A Problems/Problem 25

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