2004 AMC 12B Problems/Problem 16


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 $f(z) = z$?

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


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

We start the same as the above solution: Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$. Since we are given $|z| = 5$, this implies that $a^2+b^2=25$. We recognize the Pythagorean triple $3,4,5$ so we see that $(a,b)=(3,4)$ or $(4,3)$. So the answer is $2 \Rightarrow \mathrm{(C)}$.

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

Thanks for clearing that up IceMatrix!

Solution 3

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.

See also

2004 AMC 12B (ProblemsAnswer KeyResources)
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

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