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

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Solution by franzliszt
 
Solution by franzliszt
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===Solution 3===
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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.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:57, 16 May 2021

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

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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