Difference between revisions of "2013 AMC 12A Problems/Problem 25"
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Let <math>f : \mathbb{C} \to \mathbb{C} </math> be defined by <math> f(z) = z^2 + iz + 1 </math>. How many complex numbers <math>z </math> are there such that <math> \text{Im}(z) > 0 </math> and both the real and the imaginary parts of <math>f(z)</math> are integers with absolute value at most <math> 10 </math>? | Let <math>f : \mathbb{C} \to \mathbb{C} </math> be defined by <math> f(z) = z^2 + iz + 1 </math>. How many complex numbers <math>z </math> are there such that <math> \text{Im}(z) > 0 </math> and both the real and the imaginary parts of <math>f(z)</math> are integers with absolute value at most <math> 10 </math>? | ||
− | <math> \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D | + | <math> \textbf{(A)} \ 399 \qquad \textbf{(B)} \ 401 \qquad \textbf{(C)} \ 413 \qquad \textbf{(D)} \ 431 \qquad \textbf{(E)} \ 441 </math> |
==Solution== | ==Solution== |
Revision as of 11:39, 26 February 2013
Problem
Let be defined by
. How many complex numbers
are there such that
and both the real and the imaginary parts of
are integers with absolute value at most
?
Solution
Suppose . We look for
with
such that
are integers where
.
First, use the quadratic formula:
Generally, consider the imaginary part of a radical of a complex number: , where
.
.
Now let , then
,
,
.
Note that if and only if
. The latter is true only when we take the positive sign, and that
,
or ,
, or
.
In other words, for all ,
satisfies
, and there is one and only one
that makes it true. Therefore we are just going to count the number of ordered pairs
such that
,
are integers of magnitude no greater than
, and that
.
When , there is no restriction on
so there are
pairs;
when , there are
pairs.
So there are in total.
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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 |