Difference between revisions of "2004 AMC 12B Problems/Problem 16"

Latest revision as of 11:59, 30 March 2023

Problem

A function $f$ is defined by $f(z) = i\overline{z}$ , where $i=\sqrt{-1}$ and $\overline{z}$ is the complex conjugate of $z$ . How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$ ?

$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 2 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 8$

Solutions

Solution 1

Let $z = a+bi$ , so $\overline{z} = a-bi$ . By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$ , which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 \Rightarrow \mathrm{(C)}$ intersections.

Solution 2

Let $z=a+bi$ , like above. Therefore, $z = a+bi = i\overline{z} = i(a-bi) = ai+b$ . We move some terms around to get $bi-b = ai-a$ . We factor: $b(i-1) = a(i-1)$ . We divide out the common factor to see that $b = a$ . Next we put this into the definition of $|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25$ . Finally, $a = \pm\sqrt{\frac{25}{2}}$ , and $a$ has two solutions.

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 ===Solution 2===
-We start the same as the above solution: Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>. Since we are given <math>|z| = 5</math>, this implies that <math>a^2+b^2=25</math>. We recognize the Pythagorean triple <math>3,4,5</math> so we see that <math>(a,b)=(3,4)</math> or <math>(4,3)</math>. So the answer is <math>2 \Rightarrow \mathrm{(C)}</math>.
-Solution by franzliszt
-Comment by IceMatrix
-Hi franzliszt, I wanted to say first, that this isn't a criticism. I have seen much of your contributions and find you to be a rather impressive thinker. I just wanted to share some insight on your above solution. It doesn't actually work but happens to produce the correct answer by coincidence. I noticed this today as I was going through the problem with one of my students. The reason is you made an assumption that because (3,4) produces a magnitude of 5 that they must somehow satisfy the original problem. But they fail the second requirement. Namely that f(z) be equal to z. To demonstrate f(z)= i(a-bi)=b+ai as you state. But that in turn must be equal to z which is a+bi. So for (3,4) to be a solution it would need to be true that 4+3i be equal to 3+4i. However this is not true and so the solution fails. As the 3rd solution below this one notes, 'a' must actually be equal to 'b'. I hope you do not feel any embarrassment about this, you are an excellent problem solver and contributor and I have made similar type mistakes many times in my solving and teaching. I am posting this comment so that other viewers of the page can understand in the event they were confused by your solution.
-Best Regards,
-IceMatrix
-===Solution 3===
 Let <math>z=a+bi</math>, like above. Therefore, <math>z = a+bi = i\overline{z} = i(a-bi) = ai+b</math>. We move some terms around to get <math>bi-b = ai-a</math>. We factor: <math>b(i-1) = a(i-1)</math>. We divide out the common factor to see that <math>b = a</math>. Next we put this into the definition of <math>|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25</math>. Finally, <math>a = \pm\sqrt{\frac{25}{2}}</math>, and <math>a</math> has two solutions.

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