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Difference between revisions of "2012 AIME II Problems/Problem 6"
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To maximize this value, we must have that <math>z^2</math> is in the opposite direction of <math>1+2i</math>. The unit vector in the complex plane in the desired direction is <math>\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i</math>. Furthermore, we know that the magnitude of <math>z^2</math> is <math>25</math>, because the magnitude of <math>z</math> is <math>5</math>. From this information, we can find that <math>z^2 = \sqrt{5} (-5 - 10i)</math>
 
To maximize this value, we must have that <math>z^2</math> is in the opposite direction of <math>1+2i</math>. The unit vector in the complex plane in the desired direction is <math>\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i</math>. Furthermore, we know that the magnitude of <math>z^2</math> is <math>25</math>, because the magnitude of <math>z</math> is <math>5</math>. From this information, we can find that <math>z^2 = \sqrt{5} (-5 - 10i)</math>
  
−
Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = 125</math>
+
Squaring, we get <math>z^4 = 5 (25 - 100 + 100i) = -375 + 500i</math>. Finally, <math>c+d = -375 + 500 = \boxed{125}</math>
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2012|n=II|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:31, 20 March 2017

Problem 6

Let $z=a+bi$ be the complex number with $\vert z \vert = 5$ and $b > 0$ such that the distance between $(1+2i)z^3$ and $z^5$ is maximized, and let $z^4 = c+di$. Find $c+d$.

Solution

Let's consider the maximization constraint first: we want to maximize the value of $|z^5 - (1+2i)z^3|$ Simplifying, we have

$|z^3| * |z^2 - (1+2i)|$

$=|z|^3 * |z^2 - (1+2i)|$

$=125|z^2 - (1+2i)|$

Thus we only need to maximize the value of $|z^2 - (1+2i)|$.

To maximize this value, we must have that $z^2$ is in the opposite direction of $1+2i$. The unit vector in the complex plane in the desired direction is $\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i$. Furthermore, we know that the magnitude of $z^2$ is $25$, because the magnitude of $z$ is $5$. From this information, we can find that $z^2 = \sqrt{5} (-5 - 10i)$

Squaring, we get $z^4 = 5 (25 - 100 + 100i) = -375 + 500i$. Finally, $c+d = -375 + 500 = \boxed{125}$

See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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