Difference between revisions of "2021 AMC 12A Problems/Problem 13"
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MRENTHUSIASM (talk | contribs) m (→Solution 3 (Binomial Theorem): Shortened and generalized the solution.) |
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\end{align*}</cmath> | \end{align*}</cmath> | ||
We find the real parts of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)}</math> directly: | We find the real parts of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)}</math> directly: | ||
+ | |||
* For <math>\textbf{(B)},</math> we have <math>\mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math> | * For <math>\textbf{(B)},</math> we have <math>\mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math> | ||
+ | |||
* For <math>\textbf{(C)},</math> we have <math>\mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math> | * For <math>\textbf{(C)},</math> we have <math>\mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math> | ||
+ | |||
* For <math>\textbf{(D)},</math> we have <math>\mathrm{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</math> | * For <math>\textbf{(D)},</math> we have <math>\mathrm{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</math> | ||
+ | |||
Therefore, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> | Therefore, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> | ||
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&=\left(a^5-10a^3b^2+5ab^4\right) + \left(5a^4b-10a^2b^3+b^5\right)i. | &=\left(a^5-10a^3b^2+5ab^4\right) + \left(5a^4b-10a^2b^3+b^5\right)i. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | We find the full expansions of <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)},</math> then extract their real parts: | ||
− | * For <math>\textbf{(B)},</math> we have < | + | * For <math>\textbf{(B)},</math> we have <math>\left(-\sqrt3+i\right)^5=16\sqrt3+16i,</math> from which <math>\mathrm{Re}\left(\left(-\sqrt3+i\right)^5\right)=16\sqrt3.</math> |
− | |||
− | |||
− | |||
− | |||
− | |||
− | * For <math>\textbf{(C)},</math> we have < | + | * For <math>\textbf{(C)},</math> we have <math>\left(-\sqrt2+\sqrt2 i\right)^5=16\sqrt2-16\sqrt2i,</math> from which <math>\mathrm{Re}\left(\left(-\sqrt2+\sqrt2 i\right)^5\right)=16\sqrt2.</math> |
− | \left(-\sqrt2+\sqrt2 i\right)^5 | ||
− | |||
− | |||
− | |||
− | |||
− | * For <math>\textbf{(D)},</math> we have < | + | * For <math>\textbf{(D)},</math> we have <math>\left(-1+\sqrt3i\right)^5=-16-16\sqrt3i,</math> from which <math>\mathrm{Re}\left(\left(-1+\sqrt3 i\right)^5\right)=-16.</math> |
− | \left(-1+\sqrt3i\right)^5 | ||
− | |||
− | |||
− | |||
− | |||
Therefore, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> | Therefore, the answer is <math>\boxed{\textbf{(B) }-\sqrt3+i}.</math> |
Revision as of 17:14, 2 July 2021
Contents
- 1 Problem
- 2 Solution 1 (De Moivre's Theorem: Degrees)
- 3 Solution 2 (De Moivre's Theorem: Radians)
- 4 Solution 3 (Binomial Theorem)
- 5 Video Solution by Punxsutawney Phil
- 6 Video Solution by Hawk Math
- 7 Video Solution by OmegaLearn (Using Polar Form and De Moivre's Theorem)
- 8 Video Solution by TheBeautyofMath
- 9 See Also
Problem
Of the following complex numbers , which one has the property that
has the greatest real part?
Solution 1 (De Moivre's Theorem: Degrees)
First, .
Taking the real part of the 5th power of each we have:
,
which is negative
which is zero
Thus, the answer is .
~JHawk0224
Solution 2 (De Moivre's Theorem: Radians)
We rewrite each answer choice to the polar form where
is the magnitude of
such that
and
is the argument of
such that
By De Moivre's Theorem, the real part of is
We construct a table as follows:
Clearly, the answer is
~MRENTHUSIASM
Solution 3 (Binomial Theorem)
We evaluate the fifth power of each answer choice:
- For
we have
from which
- For
we have
from which
We will apply the Binomial Theorem to each of and
More generally, let for some real numbers
and
Two solutions follow from here:
Solution 3.1 (Real Parts Only)
To find the real part of we only need the terms with even powers of
We find the real parts of
and
directly:
- For
we have
- For
we have
- For
we have
Therefore, the answer is
~MRENTHUSIASM
Solution 3.2 (Full Expansions)
The full expansion of is
We find the full expansions of
and
then extract their real parts:
- For
we have
from which
- For
we have
from which
- For
we have
from which
Therefore, the answer is
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=FD9BE7hpRvg
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Polar Form and De Moivre's Theorem)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/ySWSHyY9TwI?t=568
~IceMatrix
See Also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.