Difference between revisions of "1992 AHSME Problems/Problem 15"

Latest revision as of 02:46, 20 February 2018

Problem

Let $i=\sqrt{-1}$ . Define a sequence of complex numbers by

$z_1=0,\quad z_{n+1}=z_{n}^2+i \text{ for } n\ge1.$ In the complex plane, how far from the origin is $z_{111}$ ?

$\text{(A) } 1\quad \text{(B) } \sqrt{2}\quad \text{(C) } \sqrt{3}\quad \text{(D) } \sqrt{110}\quad \text{(E) } \sqrt{2^{55}}$

Solution

$\fbox{B}$ Write out some terms: $0, i, -1+i, -i, -1+i, -i, -1+i, -i$ , etc., and it keeps alternating between $-1+i$ and $-i$ , so as $111$ is odd, $z_{111}$ is $-1+i$ . Thus its distance from the origin is $\sqrt{(-1)^2+1^2} = \sqrt{2}$ .

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 == Solution ==
-<math>\fbox{B}</math>
+<math>\fbox{B}</math> Write out some terms: <math>0, i, -1+i, -i, -1+i, -i, -1+i, -i</math>, etc., and it keeps alternating between <math>-1+i</math> and <math>-i</math>, so as <math>111</math> is odd, <math>z_{111}</math> is <math>-1+i</math>. Thus its distance from the origin is <math>\sqrt{(-1)^2+1^2} = \sqrt{2}</math>.
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Difference between revisions of "1992 AHSME Problems/Problem 15"

Latest revision as of 02:46, 20 February 2018

Problem

Solution

See also