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Difference between revisions of "2013 AMC 12A Problems/Problem 25"
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Suppose <math>f(z)=z^2+iz+1=c=a+bi</math>. We look for <math>z</math> with <math>Im(z)>0</math> such that <math>a,b</math> are integers where <math>|a|, |b|\leq 10</math>.
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Suppose <math>f(z)=z^2+iz+1=c=a+bi</math>. We look for <math>z</math> with <math>\text{Im}(z)>0</math> such that <math>a,b</math> are integers where <math>|a|, |b|\leq 10</math>.
  
 
First, use the quadratic formula:
 
First, use the quadratic formula:

Revision as of 13:41, 18 February 2013

Suppose $f(z)=z^2+iz+1=c=a+bi$. We look for $z$ with $\text{Im}(z)>0$ 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 $u = v+wi = r e^{i\theta}$.

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

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

Note that $Im(z)>0$ 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 $b^2 > a-1$, and there is one and only one $z$ that makes it true. Therefore we are just going to count the number of ordered pairs $(a,b)$ such that $a$, $b$ are integers of magnitude no greater than $10$, 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.

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