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Difference between revisions of "2013 AMC 12A Problems/Problem 25"

m (→‎Solution)
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==Solution==
 
==Solution==
  
−
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>.
+
Suppose <math>f(z)=z^2+iz+1=c=a+bi</math>. We look for <math>z</math> with <math>\operatorname{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:
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Generally, consider the imaginary part of a radical of a complex number: <math>\sqrt{u}</math>, where <math>u = v+wi = r e^{i\theta}</math>.
 
Generally, consider the imaginary part of a radical of a complex number: <math>\sqrt{u}</math>, where <math>u = v+wi = r e^{i\theta}</math>.
  
−
<math>Im (\sqrt{u}) = 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}}</math>.
+
<math>\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}}</math>.
  
 
Now let <math>u= -5/4 + c</math>, then <math>v = -5/4 + a</math>, <math>w=b</math>, <math>r=\sqrt{v^2 + w^2}</math>.
 
Now let <math>u= -5/4 + c</math>, then <math>v = -5/4 + a</math>, <math>w=b</math>, <math>r=\sqrt{v^2 + w^2}</math>.
  
−
Note that <math>Im(z)>0</math> if and only if <math>\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}</math>. The latter is true only when we take the positive sign, and that <math>r-v > 1/2</math>,
+
Note that <math>\operatorname{Im}(z)>0</math> if and only if <math>\pm \sqrt{\frac{r-v}{2}}>\frac{1}{2}</math>. The latter is true only when we take the positive sign, and that <math>r-v > 1/2</math>,
  
 
or <math>v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2</math>, <math>w^2 > 1/4 + v</math>, or <math>b^2 > a-1</math>.
 
or <math>v^2 + w^2 > (1/2 + v)^2 = 1/4 + v + v^2</math>, <math>w^2 > 1/4 + v</math>, or <math>b^2 > a-1</math>.

Revision as of 17:26, 15 October 2013

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.

See also

2013 AMC 12A (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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