2004 AMC 12B Problems/Problem 16

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

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

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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